GSEB Class 12 Maths Solutions Chapter 5 Continuity and Differentiability Exercise 5.2

Question 1. sin(x² + 5)
Answer: Let \( y = \sin(x^2 + 5) \).
We put \( x^2 + 5 = t \).
Therefore, \( y = \sin t \) and \( t = x^2 + 5 \).
So, \( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \cos t \cdot \frac{dt}{dx} \)
\( = \cos(x^2 + 5)\frac{d}{dx}(x^2 + 5) \)
\( = \cos(x^2 + 5) \times 2x \)
\( = 2x \cos(x^2 + 5) \).
In simple words: To differentiate this, we use the chain rule. We let the inner part be 't', differentiate sin(t), then differentiate 't' itself, and multiply these results together to get the final answer.

Exam Tip: Remember to apply the chain rule correctly when differentiating composite functions. Always identify the inner and outer functions clearly.

Question 2. cos (sin x)
Answer: Let \( y = \cos (\sin x) \).
We put \( \sin x = t \).
So, \( \frac{dy}{dt} = - \sin t \) and \( \frac{dt}{dx} = \cos x \).
Therefore, \( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = (- \sin t) \times (\cos x) \).
Substituting the value of \( t \), we get
\( \frac{dy}{dx} = - \sin (\sin x) \times \cos x \)
\( = - [\sin(\sin x)] \cos x \).
In simple words: For this function, we treat the inner sin(x) as 't'. We differentiate cos(t) to get -sin(t), then differentiate sin(x) to get cos(x), and then multiply these derivatives.

Exam Tip: Pay close attention to negative signs that arise from differentiating trigonometric functions, especially cos(x).

Question 3. sin (ax + b)
Answer: Let \( y = \sin (ax + b) \).
We put \( ax + b = t \).
Thus, \( y = \sin t \) and \( t = ax + b \).
Therefore, \( \frac{dy}{dt} = \cos t \), and \( \frac{dt}{dx} = \frac{d}{dx} (ax + b) = a \).
Now, \( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = (\cos t) \times a \)
\( = a \cos (ax + b) \).
In simple words: When differentiating sin(ax + b), we take the derivative of the sine function which gives cos(ax + b), and then we multiply this by the derivative of the inner function (ax + b), which is 'a'.

Exam Tip: For linear inner functions like (ax+b), the derivative of the inner function is simply the coefficient 'a'.

Question 4. sec(tan(\(\sqrt{x}\)))
Answer: Let \( y = \sec(\tan(\sqrt{x})) \).
We put \( \sqrt{x} = t \) and \( s = \tan t \).
This means \( y = \sec s \), \( s = \tan t \) and \( t = \sqrt{x} \).
Now, \( \frac{dy}{dx} = \frac{dy}{ds} \times \frac{ds}{dt} \times \frac{dt}{dx} \) ... (1)
So, for \( y = \sec s \), \( \frac{dy}{ds} = \sec s \tan s \).
Also, for \( s = \tan t \), \( \frac{ds}{dt} = \sec^2 t \).
Furthermore, for \( t = \sqrt{x} \), \( \frac{dt}{dx} = \frac{1}{2 \sqrt{x}} \).
Substituting these values into (1), we get
\( \frac{dy}{dx} = (\sec (\tan t) \tan (\tan t) ) \times (\sec^2 t) \times \frac{1}{2 \sqrt{x}} \).
Substituting back \( t = \sqrt{x} \) and \( s = \tan t \):
\( \frac{dy}{dx} = \frac{1}{2 \sqrt{x}} \sec(\tan \sqrt{x}) \tan(\tan \sqrt{x}) \sec^2 \sqrt{x} \).
In simple words: This problem involves a triple chain rule. We differentiate the outermost function (sec), then the middle function (tan), and finally the innermost function (square root of x), multiplying all three derivatives together.

Exam Tip: For nested functions, apply the chain rule layer by layer, starting from the outermost function and moving inwards, multiplying each derivative.

Question 5. \( \frac{\sin (a x+b)}{\cos (c x+d)} \)
Answer: Let \( y = \frac{\sin(ax+b)}{\cos(cx+d)} = \frac{u}{v} \).
Now, let \( u = \sin (ax + b) \).
Therefore, \( \frac{du}{dx} = \frac{d}{dx} \sin(ax + b) = a \cos (ax + b) \).
Further, let \( v = \cos(cx + d) \).
So, \( \frac{dv}{dx} = \frac{d}{dx} \cos(cx + d) = - \sin (cx + d) \times c = -c \sin(cx + d) \).
Now, using the quotient rule \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \):
\( \frac{dy}{dx} = \frac{\cos(cx+d) [a \cos(ax+b)] - \sin(ax+b) [-c \sin(cx+d)]}{\cos^2(cx+d)} \)
\( = \frac{a \cos(ax+b) \cos(cx+d) + c \sin(ax+b) \sin(cx+d)}{\cos^2(cx+d)} \).
In simple words: To differentiate a fraction, we use the quotient rule. This rule says to take the bottom part times the derivative of the top, minus the top part times the derivative of the bottom, all divided by the bottom part squared. We apply the chain rule for differentiating the sine and cosine terms.

Exam Tip: Clearly identify 'u' and 'v' for the quotient rule and remember to apply the chain rule to 'u' and 'v' themselves when differentiating them.

Question 6. cos x³. sin² (x5)
Answer: Let \( y = \cos x^3 \sin^2(x^5) \). We can write this as \( y = uv \),
where \( u = \cos x^3 \) and \( v = \sin^2(x^5) \).
To find \( \frac{du}{dx} \), we put \( x^3 = t \).
So, \( u = \cos t \), and \( t = x^3 \).
Therefore, \( \frac{du}{dt} = - \sin t \) and \( \frac{dt}{dx} = 3x^2 \).
By the chain rule, \( \frac{du}{dx} = \frac{du}{dt} \times \frac{dt}{dx} = (-\sin t) (3x^2) = - \sin x^3 (3x^2) = -3x^2 \sin x^3 \).
To find \( \frac{dv}{dx} \), we put \( x^5 = t \) and \( \sin t = s \).
So, \( v = s^2 \), \( s = \sin t \) and \( t = x^5 \).
Thus, \( \frac{dv}{ds} = 2s \), \( \frac{ds}{dt} = \cos t \) and \( \frac{dt}{dx} = 5x^4 \).
By the chain rule, \( \frac{dv}{dx} = \frac{dv}{ds} \times \frac{ds}{dt} \times \frac{dt}{dx} = (2s) \times (\cos t) \times (5x^4) \)
\( = 2 (\sin x^5) (\cos x^5) (5x^4) = 10x^4 \sin x^5 \cos x^5 \).
Now, using the product rule for \( y = uv \):
\( \frac{dy}{dx} = \frac{du}{dx} \times v + u \times \frac{dv}{dx} \)
\( = (-3x^2 \sin x^3) \times \sin^2 x^5 + (\cos x^3) \times (10x^4 \sin x^5 \cos x^5) \)
\( = -3x^2 \sin x^3 \sin^2 x^5 + 10x^4 \cos x^3 \sin x^5 \cos x^5 \).
We can factor out \( x^2 \sin x^5 \):
\( = x^2 \sin x^5 (-3 \sin x^3 \sin x^5 + 10x^2 \cos x^3 \cos x^5) \).
In simple words: This problem uses the product rule for two functions multiplied together, where each function itself requires the chain rule for its derivative. We find the derivative of each part separately using the chain rule, then combine them using the product rule formula.

Exam Tip: When combining product and chain rules, organize your work by finding the derivative of each component function (u' and v') first, and then substitute them into the product rule formula.

Question 7. \( \sqrt{\cot (x^{2})} \)
Answer: Let \( y = \sqrt{\cot x^2} \).
We put \( x^2 = t \) and \( \cot t = s \).
Therefore, \( y = \sqrt{s} = s^{1/2} \).
So, \( \frac{dy}{ds} = \frac{1}{2} s^{(1/2)-1} = \frac{1}{2} s^{-1/2} = \frac{1}{2\sqrt{s}} \).
Also, \( \frac{ds}{dt} = - \operatorname{cosec}^2 t \) and \( \frac{dt}{dx} = 2x \).
By the chain rule, \( \frac{dy}{dx} = \frac{dy}{ds} \times \frac{ds}{dt} \times \frac{dt}{dx} \)
\( = \left( \frac{1}{2\sqrt{s}} \right) \times (-\operatorname{cosec}^2 t) \times (2x) \).
Substituting back \( s = \cot t \) and \( t = x^2 \):
\( = \left( \frac{1}{2\sqrt{\cot x^2}} \right) \times (-\operatorname{cosec}^2 x^2) \times (2x) \)
\( = \frac{-2x \operatorname{cosec}^2 x^2}{2\sqrt{\cot x^2}} \)
\( = \frac{-x \operatorname{cosec}^2 x^2}{\sqrt{\cot x^2}} \).
In simple words: We have a function inside a function inside another function. We differentiate the square root, then the cotangent, and finally \( x^2 \), multiplying the results together and substituting back the original variables.

Exam Tip: Remember that \( \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}} \). Be careful with the derivatives of trigonometric functions and their nested arguments.

Question 8. cos \(\sqrt{x}\)
Answer: Let \( y = \cos \sqrt{x} \).
We put \( \sqrt{x} = t \).
Therefore, \( y = \cos t \) and \( t = \sqrt{x} \).
So, \( \frac{dy}{dt} = - \sin t \) and \( \frac{dt}{dx} = \frac{1}{2} x^{(1/2)-1} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \).
By the chain rule, \( \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} \)
\( = (- \sin t) \times \frac{1}{2\sqrt{x}} \).
Substituting back \( t = \sqrt{x} \):
\( = (- \sin \sqrt{x}) \times \frac{1}{2\sqrt{x}} \)
\( = \frac{- \sin \sqrt{x}}{2\sqrt{x}} \).
In simple words: To find the derivative of cos(\(\sqrt{x}\)), we first differentiate the cosine part to get -sin(\(\sqrt{x}\)), then we multiply this by the derivative of \(\sqrt{x}\), which is \( \frac{1}{2\sqrt{x}} \).

Exam Tip: A common mistake is forgetting to differentiate the inner function (here, \(\sqrt{x}\)). Always apply the chain rule when a function is composed of another function.

Question 9. The function f is given by \( f(x) = |x-1|, x \in R \). Show that it is not differentiable at \( x = 1 \).
Answer: The given function may be written as:
\( f(x) = \begin{cases} x-1, & \text{if } x \ge 1 \\ 1-x, & \text{if } x < 1 \end{cases} \).
We evaluate the right-hand derivative (R.H.D.) at \( x = 1 \):
R.H.D. \( = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \)
\( = \lim_{h \to 0} \frac{[(1+h)-1] - (1-1)}{h} \) (Since for \( x \ge 1 \), \( f(x) = x-1 \))
\( = \lim_{h \to 0} \frac{h}{h} = 1 \).
Now, we evaluate the left-hand derivative (L.H.D.) at \( x = 1 \):
L.H.D. \( = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} \)
\( = \lim_{h \to 0} \frac{[1-(1-h)] - (1-1)}{-h} \) (Since for \( x < 1 \), \( f(x) = 1-x \))
\( = \lim_{h \to 0} \frac{h}{-h} = -1 \).
Since R.H.D. \( \ne \) L.H.D. at \( x = 1 \), the function \( f \) is not differentiable at \( x = 1 \).
In simple words: For a function to be differentiable at a point, its derivative from the left side must match its derivative from the right side. For \( |x-1| \) at \( x=1 \), the right-hand derivative is 1, but the left-hand derivative is -1. Because these are different, the function cannot be differentiated at \( x=1 \).

Exam Tip: Absolute value functions typically have sharp corners at the point where the expression inside the absolute value becomes zero, leading to non-differentiability at that point. Always check left and right-hand derivatives.

Question 10. Prove that the greatest integer function defined by \( f(x) = [x], 0 < x < 3 \) is not differentiable at \( x = 1 \) and \( x = 3 \).
Answer:
(i) At \( x = 1 \):
We evaluate the right-hand derivative (R.H.D.) at \( x = 1 \):
R.H.D. \( = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \)
\( = \lim_{h \to 0} \frac{[1+h] - [1]}{h} \) (For a small \( h > 0 \), \( [1+h] = 1 \), and \( [1] = 1 \))
\( = \lim_{h \to 0} \frac{1-1}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \).
Now, we evaluate the left-hand derivative (L.H.D.) at \( x = 1 \):
L.H.D. \( = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} \)
\( = \lim_{h \to 0} \frac{[1-h] - [1]}{-h} \) (For a small \( h > 0 \), \( [1-h] = 0 \), and \( [1] = 1 \))
\( = \lim_{h \to 0} \frac{0-1}{-h} = \lim_{h \to 0} \frac{-1}{-h} = \lim_{h \to 0} \frac{1}{h} \). This limit does not exist (it approaches \( \infty \)). So, it is "Not defined".
Since R.H.D. \( \ne \) L.H.D. at \( x = 1 \), the function \( f \) is not differentiable at \( x = 1 \).
(ii) At \( x = 3 \):
We evaluate the right-hand derivative (R.H.D.) at \( x = 3 \):
R.H.D. \( = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h} \)
\( = \lim_{h \to 0} \frac{[3+h] - [3]}{h} \) (For a small \( h > 0 \), \( [3+h] = 3 \), and \( [3] = 3 \))
\( = \lim_{h \to 0} \frac{3-3}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \).
Now, we evaluate the left-hand derivative (L.H.D.) at \( x = 3 \):
L.H.D. \( = \lim_{h \to 0} \frac{f(3-h) - f(3)}{-h} \)
\( = \lim_{h \to 0} \frac{[3-h] - [3]}{-h} \) (For a small \( h > 0 \), \( [3-h] = 2 \), and \( [3] = 3 \))
\( = \lim_{h \to 0} \frac{2-3}{-h} = \lim_{h \to 0} \frac{-1}{-h} = \lim_{h \to 0} \frac{1}{h} \). This limit does not exist (it approaches \( \infty \)). So, it is "Not defined".
Since R.H.D. \( \ne \) L.H.D. at \( x = 3 \), the function \( f \) is not differentiable at \( x = 3 \).
In simple words: The greatest integer function, also called the floor function, jumps at integer values. Because there are sudden jumps, the slope cannot be properly defined at these points. This means the left-hand and right-hand derivatives are not equal at integers like 1 and 3, proving it's not differentiable there.

Exam Tip: The greatest integer function is discontinuous at all integer points. A function must be continuous to be differentiable, so it is never differentiable at integer points. This is a common property to remember.