RBSE Solutions Class 12 Maths Chapter 7 अवकलन Exercise 7.1

Question 1. \( \sin (x^2) \)
Answer: Let \( y = \sin (x^2) \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (\sin (x^2)) \)
We use the chain rule: \( \frac{d}{dx} (\sin u) = \cos u \cdot \frac{du}{dx} \). Here, \( u = x^2 \).
\( = \cos (x^2) \frac{d}{dx} (x^2) \)
\( = \cos (x^2) (2x) \)
\( = 2x \cos (x^2) \)
In simple words: To find the derivative, we first differentiate the 'sin' part, keeping the inside term the same. Then, we multiply that by the derivative of the inside term \( (x^2) \).

🎯 Exam Tip: Remember to apply the chain rule correctly: differentiate the outer function first, then multiply by the derivative of the inner function.

Question 2.
Answer: Let \( y = \tan (2x+3) \). (The question likely asks to differentiate this function, based on the solution steps provided).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (\tan (2x+3)) \)
Using the chain rule, where \( u = 2x+3 \): \( \frac{d}{dx} (\tan u) = \sec^2 u \cdot \frac{du}{dx} \).
\( = \sec^2 (2x+3) \frac{d}{dx}(2x+3) \)
\( = \sec^2 (2x+3) (2 \times 1 + 0) \)
\( = 2 \sec^2 (2x+3) \)
In simple words: We find the derivative of \( \tan(2x+3) \). First, we differentiate 'tan' to get 'sec-squared', keeping \( (2x+3) \) as is. Then, we multiply this by the derivative of \( (2x+3) \), which is 2.

🎯 Exam Tip: When differentiating a trigonometric function with an expression inside, always use the chain rule. Remember that the derivative of \( \tan x \) is \( \sec^2 x \).

Question 3. \( \sin (\cos (x^2)) \)
Answer: Let \( y = \sin (\cos (x^2)) \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (\sin (\cos (x^2))) \)
First, differentiate \( \sin(\text{something}) \) to get \( \cos(\text{something}) \), then multiply by the derivative of the 'something' (chain rule).
\( = \cos (\cos (x^2)) \frac{d}{dx} (\cos (x^2)) \)
Now, differentiate \( \cos(x^2) \). This also uses the chain rule: \( \frac{d}{dx} (\cos u) = -\sin u \cdot \frac{du}{dx} \). Here, \( u = x^2 \).
\( = \cos (\cos (x^2)) (-\sin (x^2)) \frac{d}{dx} (x^2) \)
\( = \cos (\cos (x^2)) (-\sin (x^2)) (2x) \)
\( = -2x \sin (x^2) \cos (\cos (x^2)) \)
In simple words: This is like peeling an onion. First, differentiate the 'sin' layer, then the 'cos' layer, and finally the inner \( x^2 \) layer, multiplying all the derivatives together.

🎯 Exam Tip: For nested functions, apply the chain rule multiple times from the outermost function to the innermost. Keep track of negative signs from derivatives like \( \cos x \rightarrow -\sin x \).

Question 4. \( \frac{\sec x - 1}{\sec x + 1} \)
Answer: Let \( y = \frac{\sec x - 1}{\sec x + 1} \).
We can simplify the expression first by writing \( \sec x = \frac{1}{\cos x} \):
\( y = \frac{\frac{1}{\cos x} - 1}{\frac{1}{\cos x} + 1} \)
Multiply numerator and denominator by \( \cos x \):
\( y = \frac{1 - \cos x}{1 + \cos x} \)
This simplified form is often easier to differentiate.
(Note: The differentiation steps for this simplified form are not provided in the source.)
In simple words: We start with a fraction involving 'sec x'. We change 'sec x' into '1 over cos x'. Then, we simplify the whole fraction to make it easier to work with, getting a new fraction with 'cos x'.

🎯 Exam Tip: Always look for opportunities to simplify trigonometric expressions before differentiating. Converting to sine and cosine is a good strategy.

Question 5. \( \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \)
Answer: Let \( y = \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \).
To simplify, we multiply the numerator and denominator by the conjugate of the denominator, which is \( (\sqrt{1+x}-\sqrt{1-x}) \):
\( y = \frac{(\sqrt{1+x}-\sqrt{1-x})}{(\sqrt{1+x}+\sqrt{1-x})} \times \frac{(\sqrt{1+x}-\sqrt{1-x})}{(\sqrt{1+x}-\sqrt{1-x})} \)
\( = \frac{(\sqrt{1+x}-\sqrt{1-x})^2}{(\sqrt{1+x})^2-(\sqrt{1-x})^2} \)
Using \( (a-b)^2 = a^2-2ab+b^2 \) for the numerator and \( (a-b)(a+b) = a^2-b^2 \) for the denominator:
\( = \frac{(1+x)+(1-x)-2\sqrt{(1+x)(1-x)}}{(1+x)-(1-x)} \)
\( = \frac{1+x+1-x-2\sqrt{1-x^2}}{1+x-1+x} \)
\( = \frac{2-2\sqrt{1-x^2}}{2x} \)
\( = \frac{2(1-\sqrt{1-x^2})}{2x} \)
\( = \frac{1-\sqrt{1-x^2}}{x} \)
(Note: The differentiation steps for this simplified form are not provided in the source.)
In simple words: We have a complex fraction with square roots. We make it simpler by multiplying the top and bottom by a special term (the conjugate). This helps us get rid of the roots in the bottom and simplify the whole fraction.

🎯 Exam Tip: For expressions involving square roots in the denominator of fractions, rationalizing the denominator by multiplying by the conjugate often simplifies the expression significantly before differentiation.

Question 6. \( \sin x^{\circ} \)
Answer: Given \( y = \sin x^{\circ} \).
Before differentiating, we must convert degrees to radians because calculus formulas are based on radians. The conversion is \( 1^{\circ} = \frac{\pi}{180} \) radians. So, \( x^{\circ} = \frac{\pi x}{180} \) radians.
Thus, \( y = \sin \left(\frac{\pi x}{180}\right) \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} \left(\sin \left(\frac{\pi x}{180}\right)\right) \)
Applying the chain rule: \( \frac{d}{dx} (\sin u) = \cos u \cdot \frac{du}{dx} \). Here, \( u = \frac{\pi x}{180} \).
\( = \cos \left(\frac{\pi x}{180}\right) \frac{d}{dx} \left(\frac{\pi x}{180}\right) \)
\( = \cos \left(\frac{\pi x}{180}\right) \left(\frac{\pi}{180}\right) \)
\( = \frac{\pi}{180} \cos \left(\frac{\pi x}{180}\right) \)
This can also be written back in terms of degrees:
\( = \frac{\pi}{180} \cos x^{\circ} \)
In simple words: When differentiating a trig function with degrees, first change the degrees to radians. Then use the chain rule to differentiate, just like with any other function. The constant \( \pi/180 \) comes out from the inner derivative.

🎯 Exam Tip: Always convert angles from degrees to radians before performing any calculus operations like differentiation or integration. This is a crucial first step for trigonometric functions.

Question 7. \( \log_e \sqrt{\frac{1-\cos x}{1+\cos x}} \)
Answer: Let \( y = \log_e \sqrt{\frac{1-\cos x}{1+\cos x}} \).
We can simplify this expression using logarithm properties: \( \log_e (A^B) = B \log_e A \) and \( \log_e (\frac{A}{B}) = \log_e A - \log_e B \).
\( y = \log_e \left(\frac{1-\cos x}{1+\cos x}\right)^{1/2} \)
\( = \frac{1}{2} \log_e \left(\frac{1-\cos x}{1+\cos x}\right) \)
\( = \frac{1}{2} [\log_e (1-\cos x) - \log_e (1+\cos x)] \)
Now, differentiate with respect to \( x \):
\( \frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx} (\log_e (1-\cos x)) - \frac{d}{dx} (\log_e (1+\cos x)) \right] \)
Using \( \frac{d}{dx} (\log_e u) = \frac{1}{u} \cdot \frac{du}{dx} \):
\( = \frac{1}{2} \left[ \frac{1}{1-\cos x} \frac{d}{dx} (1-\cos x) - \frac{1}{1+\cos x} \frac{d}{dx} (1+\cos x) \right] \)
\( = \frac{1}{2} \left[ \frac{1}{1-\cos x} (0-(-\sin x)) - \frac{1}{1+\cos x} (0+(-\sin x)) \right] \)
\( = \frac{1}{2} \left[ \frac{\sin x}{1-\cos x} - \frac{-\sin x}{1+\cos x} \right] \)
\( = \frac{1}{2} \left[ \frac{\sin x}{1-\cos x} + \frac{\sin x}{1+\cos x} \right] \)
Find a common denominator:
\( = \frac{1}{2} \left[ \frac{\sin x (1+\cos x) + \sin x (1-\cos x)}{(1-\cos x)(1+\cos x)} \right] \)
\( = \frac{1}{2} \left[ \frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{1-\cos^2 x} \right] \)
\( = \frac{1}{2} \left[ \frac{2 \sin x}{\sin^2 x} \right] \)
\( = \frac{1}{2} \left[ \frac{2}{\sin x} \right] \)
\( = \frac{1}{\sin x} \)
\( = \operatorname{cosec} x \)
In simple words: We simplify the logarithm first by bringing the square root out as a \( 1/2 \) factor and splitting the division into subtraction. Then we differentiate each log term, remembering the chain rule. Finally, we combine the fractions and simplify to get the answer.

🎯 Exam Tip: Always use logarithm properties to simplify expressions before differentiating. This can turn a complicated quotient rule problem into simpler derivatives of sums/differences, making it much easier to solve.

Question 9. \( \log \left(\frac{1+\sin x}{1-\sin x}\right) \)
Answer: Let \( y = \log \left(\frac{1+\sin x}{1-\sin x}\right) \).
Using logarithm property \( \log (\frac{A}{B}) = \log A - \log B \):
\( y = \log (1+\sin x) - \log (1-\sin x) \)
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (\log (1+\sin x)) - \frac{d}{dx} (\log (1-\sin x)) \)
Using \( \frac{d}{dx} (\log u) = \frac{1}{u} \cdot \frac{du}{dx} \):
\( = \frac{1}{1+\sin x} \frac{d}{dx} (1+\sin x) - \frac{1}{1-\sin x} \frac{d}{dx} (1-\sin x) \)
\( = \frac{1}{1+\sin x} (\cos x) - \frac{1}{1-\sin x} (-\cos x) \)
\( = \frac{\cos x}{1+\sin x} + \frac{\cos x}{1-\sin x} \)
Find a common denominator:
\( = \frac{\cos x (1-\sin x) + \cos x (1+\sin x)}{(1+\sin x)(1-\sin x)} \)
\( = \frac{\cos x - \sin x \cos x + \cos x + \sin x \cos x}{1-\sin^2 x} \)
\( = \frac{2 \cos x}{\cos^2 x} \)
\( = \frac{2}{\cos x} \)
\( = 2 \sec x \)
In simple words: We first use log rules to split the complex fraction into two simpler log terms. Then we differentiate each term separately, remembering the chain rule for the inner part. Finally, we combine the resulting fractions and simplify them.

🎯 Exam Tip: When differentiating logarithms of quotients, it is almost always easier to use the logarithm property \( \log(A/B) = \log A - \log B \) first, rather than applying the quotient rule directly to the original expression.

Question 10. \( \log_e \left(\frac{x+\sqrt{x^2+a^2}}{a}\right) \)
Answer: Let \( y = \log_e \left(\frac{x+\sqrt{x^2+a^2}}{a}\right) \).
Using logarithm property \( \log_e (\frac{A}{B}) = \log_e A - \log_e B \):
\( y = \log_e (x+\sqrt{x^2+a^2}) - \log_e a \)
Now, differentiate with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (\log_e (x+\sqrt{x^2+a^2})) - \frac{d}{dx} (\log_e a) \)
The derivative of \( \log_e a \) (where \( a \) is a constant) is 0.
For the first term, using \( \frac{d}{dx} (\log_e u) = \frac{1}{u} \cdot \frac{du}{dx} \):
\( = \frac{1}{x+\sqrt{x^2+a^2}} \frac{d}{dx} (x+\sqrt{x^2+a^2}) - 0 \)
\( = \frac{1}{x+\sqrt{x^2+a^2}} \left( 1 + \frac{d}{dx} (\sqrt{x^2+a^2}) \right) \)
To differentiate \( \sqrt{x^2+a^2} \), use the chain rule: \( \frac{d}{dx} (\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \). Here, \( u = x^2+a^2 \).
\( \frac{d}{dx} (\sqrt{x^2+a^2}) = \frac{1}{2\sqrt{x^2+a^2}} \frac{d}{dx} (x^2+a^2) \)
\( = \frac{1}{2\sqrt{x^2+a^2}} (2x+0) \)
\( = \frac{2x}{2\sqrt{x^2+a^2}} = \frac{x}{\sqrt{x^2+a^2}} \)
Substitute this back:
\( \frac{dy}{dx} = \frac{1}{x+\sqrt{x^2+a^2}} \left( 1 + \frac{x}{\sqrt{x^2+a^2}} \right) \)
\( = \frac{1}{x+\sqrt{x^2+a^2}} \left( \frac{\sqrt{x^2+a^2} + x}{\sqrt{x^2+a^2}} \right) \)
Notice that \( (x+\sqrt{x^2+a^2}) \) in the denominator cancels with \( (\sqrt{x^2+a^2} + x) \) in the numerator.
\( = \frac{1}{\sqrt{x^2+a^2}} \)
In simple words: First, we use log rules to separate the fraction into two log terms. The log of a constant \( (a) \) becomes zero when differentiated. For the other log term, we use the chain rule. Differentiating the square root part also involves the chain rule. Finally, we simplify the whole expression.

🎯 Exam Tip: Always simplify logarithmic expressions using properties before differentiating. Be careful with the chain rule when dealing with nested functions like \( \sqrt{x^2+a^2} \), differentiating from outside in.

Question 11. \( \log_e \left(\frac{x^2+x+1}{x^2-x+1}\right) \)
Answer: Let \( y = \log_e \left(\frac{x^2+x+1}{x^2-x+1}\right) \).
Using logarithm property \( \log_e (\frac{A}{B}) = \log_e A - \log_e B \):
\( y = \log_e (x^2+x+1) - \log_e (x^2-x+1) \)
Now, differentiate with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (\log_e (x^2+x+1)) - \frac{d}{dx} (\log_e (x^2-x+1)) \)
Using \( \frac{d}{dx} (\log_e u) = \frac{1}{u} \cdot \frac{du}{dx} \):
\( = \frac{1}{x^2+x+1} \frac{d}{dx} (x^2+x+1) - \frac{1}{x^2-x+1} \frac{d}{dx} (x^2-x+1) \)
\( = \frac{1}{x^2+x+1} (2x+1) - \frac{1}{x^2-x+1} (2x-1) \)
To combine these fractions, find a common denominator:
\( = \frac{(2x+1)(x^2-x+1) - (2x-1)(x^2+x+1)}{(x^2+x+1)(x^2-x+1)} \)
Expand the numerator:
\( (2x+1)(x^2-x+1) = 2x^3 - 2x^2 + 2x + x^2 - x + 1 = 2x^3 - x^2 + x + 1 \)
\( (2x-1)(x^2+x+1) = 2x^3 + 2x^2 + 2x - x^2 - x - 1 = 2x^3 + x^2 + x - 1 \)
Numerator becomes:
\( (2x^3 - x^2 + x + 1) - (2x^3 + x^2 + x - 1) \)
\( = 2x^3 - x^2 + x + 1 - 2x^3 - x^2 - x + 1 \)
\( = -2x^2 + 2 \)
\( = 2(1-x^2) \)
Now for the denominator, use the form \( (a+b)(a-b) = a^2-b^2 \), where \( a = x^2+1 \) and \( b = x \).
\( (x^2+1+x)(x^2+1-x) = ((x^2+1)+x)((x^2+1)-x) = (x^2+1)^2 - x^2 \)
\( = (x^4+2x^2+1) - x^2 = x^4+x^2+1 \)
So, \( \frac{dy}{dx} = \frac{2(1-x^2)}{x^4+x^2+1} \)
In simple words: We first simplify the problem by separating the log of the fraction into two simpler log terms. Then we differentiate each log term using the chain rule. After that, we combine the two resulting fractions by finding a common bottom part and simplify the top part to get the final answer.

🎯 Exam Tip: Simplifying a complex logarithmic expression into a difference of logarithms before differentiation often leads to a much easier calculation. Pay close attention to algebraic simplification after differentiation.

Question 12. \( \tan (\log_e \sqrt{1+x^2}) \)
Answer: Let \( y = \tan (\log_e \sqrt{1+x^2}) \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (\tan (\log_e \sqrt{1+x^2})) \)
Apply the chain rule multiple times. First, for \( \tan u \), where \( u = \log_e \sqrt{1+x^2} \):
\( = \sec^2 (\log_e \sqrt{1+x^2}) \frac{d}{dx} (\log_e \sqrt{1+x^2}) \)
Next, for \( \log_e v \), where \( v = \sqrt{1+x^2} \):
\( = \sec^2 (\log_e \sqrt{1+x^2}) \frac{1}{\sqrt{1+x^2}} \frac{d}{dx} (\sqrt{1+x^2}) \)
Finally, for \( \sqrt{w} \), where \( w = 1+x^2 \):
\( = \sec^2 (\log_e \sqrt{1+x^2}) \frac{1}{\sqrt{1+x^2}} \frac{1}{2\sqrt{1+x^2}} \frac{d}{dx} (1+x^2) \)
\( = \sec^2 (\log_e \sqrt{1+x^2}) \frac{1}{\sqrt{1+x^2}} \frac{1}{2\sqrt{1+x^2}} (2x) \)
\( = \sec^2 (\log_e \sqrt{1+x^2}) \frac{1}{1+x^2} \frac{2x}{2} \)
\( = \sec^2 (\log_e \sqrt{1+x^2}) \frac{x}{1+x^2} \)
\( = \frac{x}{1+x^2} \sec^2 (\log_e \sqrt{1+x^2}) \)
In simple words: We have a function with three layers: tan, then log, then square root. We differentiate each layer one by one from outside to inside, multiplying all the derivatives together.

🎯 Exam Tip: When dealing with composite functions (functions inside other functions), carefully apply the chain rule step-by-step. Break down the differentiation into smaller, manageable parts.

Question 14. \( \log_e (\sec x + \tan x) \)
Answer: Let \( y = \log_e (\sec x + \tan x) \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (\log_e (\sec x + \tan x)) \)
Using \( \frac{d}{dx} (\log_e u) = \frac{1}{u} \cdot \frac{du}{dx} \):
\( = \frac{1}{\sec x + \tan x} \frac{d}{dx} (\sec x + \tan x) \)
Recall that \( \frac{d}{dx} (\sec x) = \sec x \tan x \) and \( \frac{d}{dx} (\tan x) = \sec^2 x \).
\( = \frac{1}{\sec x + \tan x} (\sec x \tan x + \sec^2 x) \)
Factor out \( \sec x \) from the second term:
\( = \frac{1}{\sec x + \tan x} (\sec x (\tan x + \sec x)) \)
Cancel the common term \( (\sec x + \tan x) \):
\( = \sec x \)
In simple words: We differentiate the log function first. Then, we multiply it by the derivative of the expression inside the log. After finding the derivatives of sec x and tan x, we simplify the terms to get the final answer.

🎯 Exam Tip: Remember the derivatives of common trigonometric functions like \( \sec x \) and \( \tan x \). Also, look for opportunities to factor and cancel terms to simplify the final result.

Question 15. \( \sin^3 x \sin 3x \)
Answer: Let \( y = \sin^3 x \sin 3x \).
We can simplify \( \sin 3x \) using the triple angle formula: \( \sin 3x = 3 \sin x - 4 \sin^3 x \).
Substitute this into the expression for \( y \):
\( y = \sin^3 x (3 \sin x - 4 \sin^3 x) \)
\( y = 3 \sin^4 x - 4 \sin^6 x \)
Now, differentiate with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (3 \sin^4 x - 4 \sin^6 x) \)
Using the chain rule \( \frac{d}{dx} (u^n) = n u^{n-1} \frac{du}{dx} \), where \( u = \sin x \) and \( \frac{du}{dx} = \cos x \):
\( = 3 \cdot 4 \sin^{4-1} x \cdot \frac{d}{dx} (\sin x) - 4 \cdot 6 \sin^{6-1} x \cdot \frac{d}{dx} (\sin x) \)
\( = 12 \sin^3 x \cos x - 24 \sin^5 x \cos x \)
Factor out common terms, \( 12 \sin^3 x \cos x \):
\( = 12 \sin^3 x \cos x (1 - 2 \sin^2 x) \)
Recall the double angle formula \( \cos 2x = 1 - 2 \sin^2 x \):
\( = 12 \sin^3 x \cos x \cos 2x \)
This can be further simplified using \( \sin 2x = 2 \sin x \cos x \):
\( = 6 \sin^2 x (2 \sin x \cos x) \cos 2x \)
\( = 6 \sin^2 x \sin 2x \cos 2x \)
Again, use \( \sin 2x = 2 \sin x \cos x \), so \( \sin 4x = 2 \sin 2x \cos 2x \):
\( = 3 \sin^2 x (2 \sin 2x \cos 2x) \)
\( = 3 \sin^2 x \sin 4x \)
In simple words: First, we use a special formula to change \( \sin 3x \) into terms of \( \sin x \). This helps to simplify the whole expression. Then, we differentiate this new, simpler expression using the power rule and chain rule. Finally, we simplify the answer using other trigonometric identities.

🎯 Exam Tip: Look for opportunities to simplify trigonometric expressions using identities (like triple angle or double angle formulas) *before* differentiating. This can make the differentiation process much more straightforward and lead to a more concise final answer.