OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Exercise 8 (E)

S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(e)

Differentiate w.r.t. x :

Question 1.
(i) \( e^{3x} \)
(ii) \( e^{\cos x} \)
(iii) \( e^{-x/2} \)
(iv) \( e^{x^2+2x} \)
(v) \( e^{\sqrt{1+x+x^2}} \)
(vi) \( e^{\sin \sqrt{x}} \)
(vii) \( e^{\frac{x^2}{1+x^2}} \)
Answer:
(i) Let \( y = e^{3x} \)
To find the derivative, we use the chain rule. The derivative of \( e^u \) is \( e^u \cdot \frac{du}{dx} \).
Here, \( u = 3x \), so \( \frac{du}{dx} = 3 \).
\[ \frac{dy}{dx} = e^{3x} \cdot \frac{d}{dx}(3x) = e^{3x} \cdot 3 = 3e^{3x} \] The chain rule helps us differentiate composite functions, like an exponential function where the exponent is another function of x.
In simple words: When you have \( e \) raised to a power that is a function of \( x \), you differentiate \( e \) to that power and then multiply by the derivative of the power itself.

🎯 Exam Tip: Remember to apply the chain rule correctly: differentiate the outer function first, then multiply by the derivative of the inner function (the exponent in this case).

Question 1.
(ii) \( e^{\cos x} \)
Answer:
Let \( y = e^{\cos x} \)
We apply the chain rule. Here, \( u = \cos x \), so \( \frac{du}{dx} = -\sin x \).
\[ \frac{dy}{dx} = e^{\cos x} \cdot \frac{d}{dx}(\cos x) = e^{\cos x} \cdot (-\sin x) = -\sin x e^{\cos x} \] The derivative of \( \cos x \) is always \( -\sin x \).
In simple words: First, you find the derivative of \( e \) to the power of \( \cos x \) (which is still \( e \) to the power of \( \cos x \)). Then, you multiply this by the derivative of \( \cos x \), which is \( -\sin x \).

🎯 Exam Tip: Be careful with the sign of the derivative of trigonometric functions. The derivative of \( \cos x \) is \( -\sin x \), not \( \sin x \).

Question 1.
(iii) \( e^{-x/2} \)
Answer:
Let \( y = e^{-x/2} \)
Using the chain rule, with \( u = -x/2 \), we find \( \frac{du}{dx} = -1/2 \).
\[ \frac{dy}{dx} = e^{-x/2} \cdot \frac{d}{dx}\left(-\frac{x}{2}\right) = e^{-x/2} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2} e^{-x/2} \] When differentiating a term like \( kx \), the derivative is just \( k \).
In simple words: Take the derivative of \( e \) to the power of \( -x/2 \) (which is \( e \) to the power of \( -x/2 \) itself). Then, multiply this by the derivative of \( -x/2 \), which is \( -1/2 \).

🎯 Exam Tip: When the exponent is a fraction with a constant, like \( -x/2 \), treat it as \( -\frac{1}{2}x \) to easily find its derivative.

Question 1.
(iv) \( e^{x^2+2x} \)
Answer:
Let \( y = e^{x^2+2x} \)
Applying the chain rule, with \( u = x^2+2x \), we get \( \frac{du}{dx} = 2x+2 \).
\[ \frac{dy}{dx} = e^{x^2+2x} \cdot \frac{d}{dx}(x^2+2x) = e^{x^2+2x} \cdot (2x+2) = 2(x+1)e^{x^2+2x} \] The derivative of \( x^n \) is \( nx^{n-1} \).
In simple words: The derivative of \( e \) to the power of \( (x^2+2x) \) is first \( e \) to the power of \( (x^2+2x) \). Then, multiply this by the derivative of the exponent \( (x^2+2x) \), which is \( (2x+2) \).

🎯 Exam Tip: Always factor out common terms, such as 2 in \( (2x+2) \), to simplify the final answer.

Question 1.
(v) \( e^{\sqrt{1+x+x^2}} \)
Answer:
Let \( y = e^{\sqrt{1+x+x^2}} \)
We use the chain rule multiple times. Let \( u = \sqrt{1+x+x^2} \). The derivative of \( e^u \) is \( e^u \frac{du}{dx} \).
Now we need \( \frac{du}{dx} \). Let \( v = 1+x+x^2 \). Then \( u = \sqrt{v} = v^{1/2} \).
\[ \frac{du}{dx} = \frac{d}{dx}(v^{1/2}) = \frac{1}{2} v^{-1/2} \cdot \frac{dv}{dx} = \frac{1}{2\sqrt{1+x+x^2}} \cdot \frac{d}{dx}(1+x+x^2) \] \[ \frac{du}{dx} = \frac{1}{2\sqrt{1+x+x^2}} \cdot (0+1+2x) = \frac{1+2x}{2\sqrt{1+x+x^2}} \] Therefore,
\[ \frac{dy}{dx} = e^{\sqrt{1+x+x^2}} \cdot \frac{1+2x}{2\sqrt{1+x+x^2}} \] The derivative of \( \sqrt{f(x)} \) is \( \frac{f'(x)}{2\sqrt{f(x)}} \).
In simple words: This problem involves nested functions. First, treat the whole square root part as the exponent of \( e \). After that, take the derivative of the square root part, which also requires another small step: differentiating the terms inside the square root.

🎯 Exam Tip: For functions with multiple layers (like \( e^{\sqrt{f(x)}} \)), apply the chain rule step-by-step from the outermost function to the innermost function.

Question 1.
(vi) \( e^{\sin \sqrt{x}} \)
Answer:
Let \( y = e^{\sin \sqrt{x}} \)
Using the chain rule, let \( u = \sin \sqrt{x} \). Then \( \frac{dy}{dx} = e^{\sin \sqrt{x}} \cdot \frac{d}{dx}(\sin \sqrt{x}) \).
To find \( \frac{d}{dx}(\sin \sqrt{x}) \), let \( v = \sqrt{x} \). Then \( \frac{d}{dx}(\sin v) = \cos v \cdot \frac{dv}{dx} \).
We know \( \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \).
So, \( \frac{d}{dx}(\sin \sqrt{x}) = \cos \sqrt{x} \cdot \frac{1}{2\sqrt{x}} \).
Therefore,
\[ \frac{dy}{dx} = e^{\sin \sqrt{x}} \cdot \cos \sqrt{x} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sin \sqrt{x}} \cos \sqrt{x}}{2\sqrt{x}} \] Each layer of a composite function needs its own derivative calculation.
In simple words: This is a three-layer problem. First, differentiate \( e \) to the power of \( \sin \sqrt{x} \). Then, multiply by the derivative of \( \sin \sqrt{x} \). For that part, multiply by the derivative of \( \sqrt{x} \).

🎯 Exam Tip: When dealing with nested trigonometric functions, apply the chain rule for each level: first the `e` part, then the `sin` part, then the `sqrt` part.

Question 1.
(vii) \( e^{\frac{x^2}{1+x^2}} \)
Answer:
Let \( y = e^{\frac{x^2}{1+x^2}} \)
We apply the chain rule. Let \( u = \frac{x^2}{1+x^2} \). Then \( \frac{dy}{dx} = e^u \cdot \frac{du}{dx} \).
To find \( \frac{du}{dx} \), we use the quotient rule: \( \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \).
Here, \( f(x) = x^2 \) and \( g(x) = 1+x^2 \).
So, \( f'(x) = 2x \) and \( g'(x) = 2x \).
\[ \frac{du}{dx} = \frac{(2x)(1+x^2) - (x^2)(2x)}{(1+x^2)^2} = \frac{2x+2x^3 - 2x^3}{(1+x^2)^2} = \frac{2x}{(1+x^2)^2} \] Therefore,
\[ \frac{dy}{dx} = e^{\frac{x^2}{1+x^2}} \cdot \frac{2x}{(1+x^2)^2} \] The quotient rule is essential for differentiating rational functions.
In simple words: First, take the derivative of \( e \) to the power of \( \frac{x^2}{1+x^2} \), which is just \( e \) to that power. Then, multiply it by the derivative of the exponent \( \frac{x^2}{1+x^2} \), which you find using the quotient rule.

🎯 Exam Tip: When the exponent is a fraction, remember to apply the quotient rule carefully for its derivative before multiplying it with the derivative of the exponential function.

Question 2. Differentiate w.r.t. x :
(i) \( 3^x \)
(ii) \( 8^{\cos x} \)
(iii) \( a^{\sin x} \)
(iv) \( a^{3x^2} \)
(v) \( 5^{\log \sin x} \)
(vi) \( 10^{10x} \)
Answer:
(i) Let \( y = 3^x \)
The derivative of \( a^x \) is \( a^x \log a \). Here, \( a=3 \).
\[ \frac{dy}{dx} = 3^x \log 3 \] The natural logarithm (log) is often written as \( \ln \).
In simple words: To find the derivative of a number raised to the power of \( x \), you keep the same number raised to \( x \) and multiply it by the natural logarithm of that number.

🎯 Exam Tip: Remember the basic formula for the derivative of \( a^x \). It is one of the fundamental differentiation rules.

Question 2.
(ii) \( 8^{\cos x} \)
Answer:
Let \( y = 8^{\cos x} \)
We use the chain rule. The derivative of \( a^u \) is \( a^u \log a \cdot \frac{du}{dx} \). Here, \( a=8 \) and \( u=\cos x \).
\[ \frac{dy}{dx} = 8^{\cos x} \log 8 \cdot \frac{d}{dx}(\cos x) = 8^{\cos x} \log 8 \cdot (-\sin x) = -\sin x \cdot 8^{\cos x} \log 8 \] The derivative of \( \cos x \) is \( -\sin x \).
In simple words: You differentiate \( 8 \) to the power of \( \cos x \) by first writing \( 8 \) to the power of \( \cos x \) multiplied by \( \log 8 \). Then, you multiply this result by the derivative of \( \cos x \), which is \( -\sin x \).

🎯 Exam Tip: Always remember to apply the chain rule when the exponent of a constant base is a function of \( x \).

Question 2.
(iii) \( a^{\sin x} \)
Answer:
Let \( y = a^{\sin x} \)
Using the chain rule, with \( u=\sin x \), the derivative is \( a^{\sin x} \log a \cdot \frac{d}{dx}(\sin x) \).
\[ \frac{dy}{dx} = a^{\sin x} \log a \cdot (\cos x) = a^{\sin x} \log a \cos x \] The derivative of \( \sin x \) is \( \cos x \).
In simple words: The derivative of \( a \) to the power of \( \sin x \) is found by taking \( a \) to the power of \( \sin x \) multiplied by \( \log a \), and then multiplying that by the derivative of \( \sin x \), which is \( \cos x \).

🎯 Exam Tip: Ensure that you correctly apply the chain rule for the exponent, differentiating \( \sin x \) to get \( \cos x \).

Question 2.
(iv) \( a^{3x^2} \)
Answer:
Let \( y = a^{3x^2} \)
Applying the chain rule, with \( u=3x^2 \), the derivative is \( a^{3x^2} \log a \cdot \frac{d}{dx}(3x^2) \).
\[ \frac{dy}{dx} = a^{3x^2} \log a \cdot (3 \cdot 2x) = a^{3x^2} \log a \cdot 6x \] The derivative of \( x^n \) is \( nx^{n-1} \).
In simple words: To find the derivative of \( a \) raised to the power of \( 3x^2 \), first write \( a \) to the power of \( 3x^2 \) and multiply by \( \log a \). Then, multiply this whole expression by the derivative of \( 3x^2 \), which is \( 6x \).

🎯 Exam Tip: Don't forget to correctly differentiate the polynomial exponent, \( 3x^2 \), resulting in \( 6x \).

Question 2.
(v) \( 5^{\log \sin x} \)
Answer:
Let \( y = 5^{\log \sin x} \)
Using the chain rule, with \( u=\log \sin x \), the derivative is \( 5^{\log \sin x} \log 5 \cdot \frac{d}{dx}(\log \sin x) \).
To find \( \frac{d}{dx}(\log \sin x) \), let \( v=\sin x \). The derivative of \( \log v \) is \( \frac{1}{v} \cdot \frac{dv}{dx} \).
So, \( \frac{d}{dx}(\log \sin x) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x \).
Therefore,
\[ \frac{dy}{dx} = 5^{\log \sin x} \log 5 \cdot \cot x \] The derivative of \( \log f(x) \) is \( \frac{f'(x)}{f(x)} \).
In simple words: This involves three layers of differentiation. First, treat the whole \( \log \sin x \) as the exponent of \( 5 \). Then, multiply by the derivative of \( \log \sin x \). For this part, you multiply by the derivative of \( \sin x \).

🎯 Exam Tip: When you have nested logarithmic and trigonometric functions in the exponent, apply the chain rule meticulously to each level.

Question 2.
(vi) \( 10^{10x} \)
Answer:
Let \( y = 10^{10x} \)
Using the chain rule, with \( u=10x \), the derivative of \( a^u \) is \( a^u \log a \cdot \frac{du}{dx} \). Here, \( a=10 \).
\[ \frac{dy}{dx} = 10^{10x} \log 10 \cdot \frac{d}{dx}(10x) = 10^{10x} \log 10 \cdot 10 \] \[ \frac{dy}{dx} = 10 \cdot 10^{10x} \log 10 \] This can also be written as \( 10^{10x+1} \log 10 \).
In simple words: First, find the derivative of \( 10 \) to the power of \( 10x \), which is \( 10 \) to the power of \( 10x \) multiplied by \( \log 10 \). Then, multiply this result by the derivative of the exponent \( 10x \), which is \( 10 \).

🎯 Exam Tip: A common mistake is to forget to differentiate the exponent, or to confuse \( \log 10 \) with \( (\log 10)^2 \). Ensure all steps are clear.

Question 3. Differentiate w.r.t. x :
(i) \( x e^{-x} \)
(ii) \( e^x \cot x \)
(iii) \( e^{ax} \sin(bx) \)
(iv) \( \frac{e^x}{x} \)
(v) \( \frac{e^x}{1+\sin x} \)
(vi) \( x e^{x^2} \)
(vii) \( e^{x \sin x} \)
(viii) \( e^{ax} \cos(bx+c) \)
(ix) \( x^2 e^x \sin x \)
(x) \( \sin(e^x \log x) \)
(xi) \( e^{ax} \cos(b \tan x) \)
(xii) \( e^{x^2} \log_{10}(2x) \)
Answer:
(i) Let \( y = x e^{-x} \)
We use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u=x \) and \( v=e^{-x} \).
So, \( u'=1 \) and \( v' = e^{-x} \cdot (-1) = -e^{-x} \).
\[ \frac{dy}{dx} = (1)e^{-x} + x(-e^{-x}) = e^{-x} - x e^{-x} = e^{-x}(1-x) \] The product rule helps differentiate a multiplication of two functions.
In simple words: To differentiate \( x \) multiplied by \( e^{-x} \), you take the derivative of \( x \) and multiply by \( e^{-x} \), then add \( x \) multiplied by the derivative of \( e^{-x} \).

🎯 Exam Tip: Remember to factor out common terms like \( e^{-x} \) at the end to simplify the expression.

Question 3.
(ii) \( e^x \cot x \)
Answer:
Let \( y = e^x \cot x \)
Using the product rule, with \( u=e^x \) and \( v=\cot x \).
So, \( u'=e^x \) and \( v'=-\text{cosec}^2 x \).
\[ \frac{dy}{dx} = e^x (\cot x) + e^x (-\text{cosec}^2 x) = e^x \cot x - e^x \text{cosec}^2 x = e^x(\cot x - \text{cosec}^2 x) \] The derivative of \( \cot x \) is \( -\text{cosec}^2 x \).
In simple words: Apply the product rule. First, differentiate \( e^x \) and multiply by \( \cot x \). Then, add \( e^x \) multiplied by the derivative of \( \cot x \), which is \( -\text{cosec}^2 x \).

🎯 Exam Tip: Remember the derivatives of trigonometric functions, especially \( \cot x \), which has a negative sign in its derivative.

Question 3.
(iii) \( e^{ax} \sin(bx) \)
Answer:
Let \( y = e^{ax} \sin(bx) \)
Using the product rule, with \( u=e^{ax} \) and \( v=\sin(bx) \).
So, \( u' = e^{ax} \cdot a = ae^{ax} \).
And \( v' = \cos(bx) \cdot b = b\cos(bx) \).
\[ \frac{dy}{dx} = (ae^{ax})\sin(bx) + e^{ax}(b\cos(bx)) = ae^{ax}\sin(bx) + be^{ax}\cos(bx) \] \[ \frac{dy}{dx} = e^{ax}[a\sin(bx) + b\cos(bx)] \] The chain rule is applied when differentiating \( e^{ax} \) and \( \sin(bx) \).
In simple words: Use the product rule for these two multiplied functions. For \( e^{ax} \), its derivative is \( ae^{ax} \). For \( \sin(bx) \), its derivative is \( b\cos(bx) \). Combine them using the product rule.

🎯 Exam Tip: Remember to apply the chain rule when differentiating \( e^{ax} \) and \( \sin(bx) \), multiplying by the coefficients \( a \) and \( b \) respectively.

Question 3.
(iv) \( \frac{e^x}{x} \)
Answer:
Let \( y = \frac{e^x}{x} \)
We use the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \). Here, \( u=e^x \) and \( v=x \).
So, \( u'=e^x \) and \( v'=1 \).
\[ \frac{dy}{dx} = \frac{(e^x)(x) - (e^x)(1)}{x^2} = \frac{xe^x - e^x}{x^2} = \frac{e^x(x-1)}{x^2} \] The quotient rule is used for fractions of functions.
In simple words: Apply the quotient rule. The top part's derivative multiplied by the bottom part, minus the top part multiplied by the bottom part's derivative, all divided by the bottom part squared.

🎯 Exam Tip: Be careful with the order of terms in the numerator of the quotient rule; it's \( u'v - uv' \), not the other way around, to avoid sign errors.

Question 3.
(v) \( \frac{e^x}{1+\sin x} \)
Answer:
Let \( y = \frac{e^x}{1+\sin x} \)
Using the quotient rule, with \( u=e^x \) and \( v=1+\sin x \).
So, \( u'=e^x \) and \( v'=\cos x \).
\[ \frac{dy}{dx} = \frac{(e^x)(1+\sin x) - (e^x)(\cos x)}{(1+\sin x)^2} = \frac{e^x(1+\sin x - \cos x)}{(1+\sin x)^2} \] The derivative of a constant is zero.
In simple words: Use the quotient rule. Differentiate \( e^x \) and multiply by \( (1+\sin x) \). Then subtract \( e^x \) multiplied by the derivative of \( (1+\sin x) \), which is \( \cos x \). Finally, divide the whole thing by \( (1+\sin x) \) squared.

🎯 Exam Tip: Ensure that you correctly differentiate the denominator \( (1+\sin x) \) as \( \cos x \), remembering that the derivative of a constant (1) is 0.

Question 3.
(vi) \( x e^{x^2} \)
Answer:
Let \( y = x e^{x^2} \)
We use the product rule, with \( u=x \) and \( v=e^{x^2} \).
So, \( u'=1 \).
And \( v' = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot 2x = 2xe^{x^2} \).
\[ \frac{dy}{dx} = (1)e^{x^2} + x(2xe^{x^2}) = e^{x^2} + 2x^2e^{x^2} = e^{x^2}(1+2x^2) \] This problem combines both the product rule and the chain rule.
In simple words: Apply the product rule. For \( x \), its derivative is \( 1 \). For \( e^{x^2} \), its derivative is \( e^{x^2} \) multiplied by \( 2x \). Combine these parts using the product rule formula.

🎯 Exam Tip: This question requires both the product rule and the chain rule. Differentiate \( e^{x^2} \) carefully using the chain rule, then combine with the product rule.

Question 3.
(vii) \( e^{x \sin x} \)
Answer:
Let \( y = e^{x \sin x} \)
Using the chain rule, with \( u=x \sin x \), we have \( \frac{dy}{dx} = e^{x \sin x} \cdot \frac{d}{dx}(x \sin x) \).
To find \( \frac{d}{dx}(x \sin x) \), we use the product rule. Let \( p=x \) and \( q=\sin x \).
So, \( p'=1 \) and \( q'=\cos x \).
\( \frac{d}{dx}(x \sin x) = (1)\sin x + x(\cos x) = \sin x + x \cos x \).
Therefore,
\[ \frac{dy}{dx} = e^{x \sin x} (\sin x + x \cos x) \] The chain rule is always applied first for exponential functions with a function as an exponent.
In simple words: First, take the derivative of \( e \) to the power of \( (x \sin x) \), which is \( e \) to that same power. Then, multiply this by the derivative of the exponent \( (x \sin x) \), which requires using the product rule.

🎯 Exam Tip: Remember that the derivative of the exponent \( x \sin x \) itself requires the product rule. Don't simply multiply derivatives of \( x \) and \( \sin x \).

Question 3.
(viii) \( e^{ax} \cos(bx+c) \)
Answer:
Let \( y = e^{ax} \cos(bx+c) \)
Using the product rule, with \( u=e^{ax} \) and \( v=\cos(bx+c) \).
So, \( u' = e^{ax} \cdot a = ae^{ax} \).
And \( v' = -\sin(bx+c) \cdot b = -b\sin(bx+c) \).
\[ \frac{dy}{dx} = (ae^{ax})\cos(bx+c) + e^{ax}(-b\sin(bx+c)) \] \[ \frac{dy}{dx} = e^{ax}[a\cos(bx+c) - b\sin(bx+c)] \] Both parts \( e^{ax} \) and \( \cos(bx+c) \) require the chain rule for their derivatives.
In simple words: Apply the product rule. The derivative of \( e^{ax} \) is \( ae^{ax} \). The derivative of \( \cos(bx+c) \) is \( -b\sin(bx+c) \). Combine these using the product rule formula and factor out \( e^{ax} \).

🎯 Exam Tip: Pay close attention to the negative sign that appears when differentiating \( \cos(bx+c) \), and ensure all constant factors from the chain rule are included.

Question 3.
(ix) \( x^2 e^x \sin x \)
Answer:
Let \( y = x^2 e^x \sin x \)
We can treat this as a product of two functions, \( u=x^2 \) and \( v=e^x \sin x \), and apply the product rule.
\( u' = 2x \).
To find \( v' = \frac{d}{dx}(e^x \sin x) \), we use the product rule again for \( e^x \sin x \):
\( \frac{d}{dx}(e^x \sin x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x) \).
Now, apply the main product rule:
\[ \frac{dy}{dx} = u'v + uv' = (2x)(e^x \sin x) + (x^2)(e^x(\sin x + \cos x)) \] \[ \frac{dy}{dx} = 2xe^x \sin x + x^2 e^x \sin x + x^2 e^x \cos x \] Factor out \( e^x \):
\[ \frac{dy}{dx} = e^x[2x \sin x + x^2 \sin x + x^2 \cos x] \] When differentiating a product of three functions, you can also use the general rule: \( (fgh)' = f'gh + fg'h + fgh' \).
In simple words: This involves applying the product rule twice. First, treat \( x^2 \) as one function and \( e^x \sin x \) as another. For the second part, you'll need to apply the product rule again to find the derivative of \( e^x \sin x \). Then, combine all parts.

🎯 Exam Tip: For a product of three functions, it's often easiest to group two functions together and apply the product rule, then apply it again to the derivative of the grouped part.

Question 3.
(x) \( \sin(e^x \log x) \)
Answer:
Let \( y = \sin(e^x \log x) \)
We use the chain rule. The derivative of \( \sin u \) is \( \cos u \cdot \frac{du}{dx} \). Here, \( u = e^x \log x \).
To find \( \frac{du}{dx} = \frac{d}{dx}(e^x \log x) \), we use the product rule. Let \( p=e^x \) and \( q=\log x \).
So, \( p'=e^x \) and \( q'=\frac{1}{x} \).
\( \frac{d}{dx}(e^x \log x) = e^x \log x + e^x \cdot \frac{1}{x} = e^x \left(\log x + \frac{1}{x}\right) \).
Therefore,
\[ \frac{dy}{dx} = \cos(e^x \log x) \cdot e^x \left(\log x + \frac{1}{x}\right) \] \[ \frac{dy}{dx} = e^x \left(\frac{1}{x} + \log x\right) \cos(e^x \log x) \] The derivative of \( \log x \) is \( 1/x \).
In simple words: First, differentiate \( \sin \) to get \( \cos \), keeping the inside expression \( (e^x \log x) \) the same. Then, multiply by the derivative of this inside expression, \( (e^x \log x) \), which needs the product rule.

🎯 Exam Tip: When applying the chain rule, always differentiate the outermost function first, keeping the inner function unchanged, then multiply by the derivative of the inner function.

Question 3.
(xi) \( e^{ax} \cos(b \tan x) \)
Answer:
Let \( y = e^{ax} \cos(b \tan x) \)
Using the product rule, with \( u=e^{ax} \) and \( v=\cos(b \tan x) \).
So, \( u' = ae^{ax} \).
To find \( v' = \frac{d}{dx}(\cos(b \tan x)) \), we use the chain rule. Let \( p=b \tan x \).
Derivative of \( \cos p \) is \( -\sin p \cdot \frac{dp}{dx} \).
\( \frac{dp}{dx} = \frac{d}{dx}(b \tan x) = b \sec^2 x \).
So, \( v' = -\sin(b \tan x) \cdot b \sec^2 x = -b \sec^2 x \sin(b \tan x) \).
Now, apply the product rule:
\[ \frac{dy}{dx} = u'v + uv' = (ae^{ax})\cos(b \tan x) + e^{ax}(-b \sec^2 x \sin(b \tan x)) \] \[ \frac{dy}{dx} = e^{ax}[a\cos(b \tan x) - b \sec^2 x \sin(b \tan x)] \] The derivative of \( \tan x \) is \( \sec^2 x \).
In simple words: This problem involves the product rule and chain rule. Differentiate \( e^{ax} \) and \( \cos(b \tan x) \) separately, remembering to use the chain rule for each. Then, combine these results using the product rule.

🎯 Exam Tip: Be mindful of the chain rule inside the cosine function. Differentiating \( b \tan x \) gives \( b \sec^2 x \), which must be multiplied by \( -\sin(b \tan x) \).

Question 3.
(xii) \( e^{x^2} \log_{10}(2x) \)
Answer:
Let \( y = e^{x^2} \log_{10}(2x) \)
Using the product rule, with \( u=e^{x^2} \) and \( v=\log_{10}(2x) \).
So, \( u' = e^{x^2} \cdot 2x = 2xe^{x^2} \).
To find \( v' = \frac{d}{dx}(\log_{10}(2x)) \), we use the change of base formula \( \log_{b} A = \frac{\log_e A}{\log_e b} \).
So, \( \log_{10}(2x) = \frac{\ln(2x)}{\ln 10} \).
\( v' = \frac{1}{\ln 10} \cdot \frac{d}{dx}(\ln(2x)) = \frac{1}{\ln 10} \cdot \frac{1}{2x} \cdot 2 = \frac{1}{x \ln 10} \).
Now, apply the product rule:
\[ \frac{dy}{dx} = u'v + uv' = (2xe^{x^2})\log_{10}(2x) + e^{x^2}\left(\frac{1}{x \ln 10}\right) \] \[ \frac{dy}{dx} = e^{x^2} \left[2x \log_{10}(2x) + \frac{1}{x \ln 10}\right] \] Remember that \( \log_{10} e = \frac{1}{\ln 10} \), so \( \frac{1}{x \ln 10} \) can be written as \( \frac{\log_{10} e}{x} \).
\[ \frac{dy}{dx} = e^{x^2} \left[2x \log_{10}(2x) + \frac{\log_{10} e}{x}\right] \] Converting \( \log_{10} \) to natural logarithm is often necessary for differentiation.
In simple words: This problem needs the product rule. First, differentiate \( e^{x^2} \) (which involves the chain rule, giving \( 2xe^{x^2} \)). Then, differentiate \( \log_{10}(2x) \) by converting it to natural logarithm first and then applying the chain rule. Finally, combine these two parts using the product rule.

🎯 Exam Tip: For logarithms with a base other than \( e \), always convert them to the natural logarithm \( \ln \) before differentiating. Remember \( \log_b x = \frac{\ln x}{\ln b} \).

Question 4. Differentiate w.r.t. x :
(i) \( \log (e^x + e^{-x}) \)
(ii) \( \log \left(\frac{e^x}{e^x+1}\right) \)
(iii) \( \frac{e^x-e^{-x}}{e^x+e^{-x}} \)
(iv) \( \log x + e^{\sqrt{x}} \)
(v) \( e^{\tan x} \times \log \tan x \)
Answer:
(i) Let \( y = \log (e^x + e^{-x}) \)
Using the chain rule, the derivative of \( \log u \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = e^x + e^{-x} \).
\[ \frac{du}{dx} = \frac{d}{dx}(e^x + e^{-x}) = e^x + e^{-x}(-1) = e^x - e^{-x} \] Therefore,
\[ \frac{dy}{dx} = \frac{1}{e^x + e^{-x}} \cdot (e^x - e^{-x}) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \] The derivative of \( e^{-x} \) is \( -e^{-x} \).
In simple words: To differentiate \( \log(e^x + e^{-x}) \), first write \( 1 \) divided by \( (e^x + e^{-x}) \). Then, multiply this by the derivative of \( (e^x + e^{-x}) \), which is \( (e^x - e^{-x}) \).

🎯 Exam Tip: Be careful with the derivative of \( e^{-x} \), which is \( -e^{-x} \) due to the chain rule for the negative exponent.

Question 4.
(ii) \( \log \left(\frac{e^x}{e^x+1}\right) \)
Answer:
Let \( y = \log \left(\frac{e^x}{e^x+1}\right) \)
We can simplify using logarithm properties: \( \log \left(\frac{A}{B}\right) = \log A - \log B \).
\[ y = \log(e^x) - \log(e^x+1) \] Since \( \log(e^x) = x \log e = x \cdot 1 = x \), we have
\[ y = x - \log(e^x+1) \] Now, differentiate with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(\log(e^x+1)) \] \[ \frac{dy}{dx} = 1 - \frac{1}{e^x+1} \cdot \frac{d}{dx}(e^x+1) = 1 - \frac{1}{e^x+1} \cdot (e^x+0) \] \[ \frac{dy}{dx} = 1 - \frac{e^x}{e^x+1} = \frac{(e^x+1) - e^x}{e^x+1} = \frac{1}{e^x+1} \] Simplifying the expression using log rules makes differentiation easier.
In simple words: First, use logarithm rules to split the expression into \( \log e^x \) (which simplifies to \( x \)) minus \( \log(e^x+1) \). Then, differentiate \( x \) (which is \( 1 \)) and differentiate \( \log(e^x+1) \) separately, then subtract the results.

🎯 Exam Tip: Always look for opportunities to simplify expressions using logarithm properties before differentiating; it often makes the process much simpler.

Question 4.
(iii) \( \frac{e^x-e^{-x}}{e^x+e^{-x}} \)
Answer:
Let \( y = \frac{e^x-e^{-x}}{e^x+e^{-x}} \)
This function is actually \( \tanh x \). The derivative of \( \tanh x \) is \( \text{sech}^2 x \).
Using the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).
Here, \( u = e^x-e^{-x} \) and \( v = e^x+e^{-x} \).
\( u' = e^x - (-e^{-x}) = e^x+e^{-x} \).
\( v' = e^x + (-e^{-x}) = e^x-e^{-x} \).
\[ \frac{dy}{dx} = \frac{(e^x+e^{-x})(e^x+e^{-x}) - (e^x-e^{-x})(e^x-e^{-x})}{(e^x+e^{-x})^2} \] \[ \frac{dy}{dx} = \frac{(e^x+e^{-x})^2 - (e^x-e^{-x})^2}{(e^x+e^{-x})^2} \] Expand the numerator using \( (A+B)^2 = A^2+B^2+2AB \) and \( (A-B)^2 = A^2+B^2-2AB \):
\( (e^x)^2 + (e^{-x})^2 + 2e^x e^{-x} = e^{2x} + e^{-2x} + 2 \).
\( (e^x)^2 + (e^{-x})^2 - 2e^x e^{-x} = e^{2x} + e^{-2x} - 2 \).
Numerator: \( (e^{2x} + e^{-2x} + 2) - (e^{2x} + e^{-2x} - 2) = 4 \).
Therefore,
\[ \frac{dy}{dx} = \frac{4}{(e^x+e^{-x})^2} \] This function is also known as the hyperbolic tangent function.
In simple words: This problem needs the quotient rule. Find the derivative of the top part and multiply it by the bottom part. Then subtract the top part multiplied by the derivative of the bottom part. All this is divided by the bottom part squared.

🎯 Exam Tip: When using the quotient rule, be very careful with algebraic expansions and simplifications, especially when dealing with terms like \( e^x \) and \( e^{-x} \).

Question 4.
(iv) \( \log x + e^{\sqrt{x}} \)
Answer:
Let \( y = \log x + e^{\sqrt{x}} \)
We differentiate each term separately.
\[ \frac{d}{dx}(\log x) = \frac{1}{x} \] For \( e^{\sqrt{x}} \), use the chain rule. Let \( u=\sqrt{x} \). Then \( \frac{d}{dx}(e^u) = e^u \cdot \frac{du}{dx} \).
\( \frac{du}{dx} = \frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).
So, \( \frac{d}{dx}(e^{\sqrt{x}}) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \).
Combining these,
\[ \frac{dy}{dx} = \frac{1}{x} + \frac{e^{\sqrt{x}}}{2\sqrt{x}} \] The derivative of a sum is the sum of the derivatives.
In simple words: Differentiate each part of the sum separately. The derivative of \( \log x \) is \( 1/x \). For \( e^{\sqrt{x}} \), first differentiate \( e \) to that power, then multiply by the derivative of \( \sqrt{x} \), which is \( 1/(2\sqrt{x}) \).

🎯 Exam Tip: When differentiating a sum or difference of terms, remember to differentiate each term independently and then combine the results.

Question 4.
(v) \( e^{\tan x} \times \log \tan x \)
Answer:
Let \( y = e^{\tan x} \log \tan x \)
We use the product rule, with \( u=e^{\tan x} \) and \( v=\log \tan x \).
To find \( u' = \frac{d}{dx}(e^{\tan x}) \), use the chain rule: \( e^{\tan x} \cdot \frac{d}{dx}(\tan x) = e^{\tan x} \sec^2 x \).
To find \( v' = \frac{d}{dx}(\log \tan x) \), use the chain rule: \( \frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x) = \frac{1}{\tan x} \sec^2 x = \cot x \sec^2 x \).
Now, apply the product rule:
\[ \frac{dy}{dx} = u'v + uv' = (e^{\tan x} \sec^2 x)(\log \tan x) + (e^{\tan x})(\cot x \sec^2 x) \] \[ \frac{dy}{dx} = e^{\tan x} \sec^2 x (\log \tan x + \cot x) \] The derivative of \( \tan x \) is \( \sec^2 x \).
In simple words: This problem requires both the product rule and chain rule. Differentiate \( e^{\tan x} \) (using chain rule) and \( \log \tan x \) (using chain rule) separately. Then, combine these results using the product rule.

🎯 Exam Tip: Be careful to apply the chain rule correctly for both \( e^{\tan x} \) and \( \log \tan x \), remembering the derivatives of \( \tan x \) and \( \log u \).

Question 5. Differentiate w.r.t. x :
\( e^{\log \left(x+\sqrt{x^2-a^2}\right)} \)
Answer:
Let \( y = e^{\log \left(x+\sqrt{x^2-a^2}\right)} \)
Using the property \( e^{\log A} = A \), we can simplify the function first:
\[ y = x+\sqrt{x^2-a^2} \] Now, differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{x^2-a^2}) \] \[ \frac{d}{dx}(x) = 1 \] For \( \frac{d}{dx}(\sqrt{x^2-a^2}) \), use the chain rule. Let \( u = x^2-a^2 \). Then \( \sqrt{u} = u^{1/2} \).
\[ \frac{d}{dx}(\sqrt{x^2-a^2}) = \frac{1}{2}(x^2-a^2)^{-1/2} \cdot \frac{d}{dx}(x^2-a^2) \] \[ = \frac{1}{2\sqrt{x^2-a^2}} \cdot (2x-0) = \frac{2x}{2\sqrt{x^2-a^2}} = \frac{x}{\sqrt{x^2-a^2}} \] Combining these results,
\[ \frac{dy}{dx} = 1 + \frac{x}{\sqrt{x^2-a^2}} \] Simplifying the function before differentiating often makes the process much simpler.
In simple words: First, simplify the expression using the rule that \( e \) raised to the power of \( \log(\text{something}) \) is just that \( \text{something} \). So, \( y \) becomes \( x + \sqrt{x^2-a^2} \). Then, differentiate \( x \) (which is \( 1 \)) and differentiate \( \sqrt{x^2-a^2} \) separately (using the chain rule), and add the results.

🎯 Exam Tip: Always look for logarithmic or exponential properties (like \( e^{\log A} = A \)) to simplify the function before differentiating, as it can significantly reduce complexity.