GSEB Class 12 Maths Solutions Chapter 5 Continuity and Differentiability Exercise 5.7

Question 1. Find the second derivative of \( x^2 + 3x + 2 \).
Answer: Let \( y = x^2 + 3x + 2 \).
First, we calculate the initial derivative:
\( \frac { dy }{ dx } = 2x + 3 \)
Next, we compute the second derivative by differentiating the first derivative:
\( \frac{d^{2} y}{d x^{2}} = \frac { d }{ dx }(2x + 3) = 2 \).
In simple words: We find the rate of change once, then find the rate of change of that result. For \( x^2 + 3x + 2 \), the first derivative is \( 2x+3 \), and the second derivative is 2.

Exam Tip: Remember to differentiate each term separately and use the power rule. The derivative of a constant is always zero.

Question 2. Find the second derivative of \( x^{20} \).
Answer: Let \( y = x^{20} \).
First, we determine the initial derivative using the power rule:
\( \frac { dy }{ dx } = 20x^{19} \)
Then, we find the second derivative by differentiating this result again:
\( \frac{d^{2} y}{d x^{2}} = \frac { d }{ dx }(20x^{19}) = 20 \times 19x^{18} = 380x^{18} \).
In simple words: To find the second derivative of \( x^{20} \), first multiply the power by the coefficient and reduce the power by one, then repeat this process with the new expression.

Exam Tip: Pay close attention to the power rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \). Apply it carefully for each differentiation step.

Question 3. Find the second derivative of \( x \cos x \).
Answer: Let \( y = x \cos x \).
To find the first derivative, we use the product rule \( (uv)' = u'v + uv' \):
\( \frac { dy }{ dx } = x \frac { d }{ dx } (\cos x) + \cos x \frac { d }{ dx }(x) \)
\( = x(-\sin x) + \cos x \cdot 1 \)
\( = -x \sin x + \cos x \)
Now, we calculate the second derivative by differentiating \( -x \sin x + \cos x \). This involves the product rule again for the first term:
\( \frac{d^{2} y}{d x^{2}} = -\frac { d }{ dx }(x \sin x) + \frac { d }{ dx } (\cos x) \)
\( = - (x \cos x + \sin x \cdot 1) - \sin x \)
\( = -x \cos x - \sin x - \sin x \)
\( = -x \cos x - 2 \sin x \).
In simple words: We apply the product rule twice. First, to get the initial derivative of \( x \cos x \), and then again to get the second derivative. Watch out for the negative signs and similar terms.

Exam Tip: The product rule is crucial for these types of questions. Double-check your differentiation of trigonometric functions, and be careful with negative signs when simplifying.

Question 4. Find the second derivative of \( \log x \).
Answer: Let \( y = \log x \).
First, we calculate the initial derivative of \( \log x \):
\( \frac { dy }{ dx } = \frac { 1 }{ x } = x^{-1} \)
Next, we compute the second derivative by differentiating \( x^{-1} \):
\( \frac{d^{2} y}{d x^{2}} = \frac { d }{ dx }(x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^{2}} \).
In simple words: The derivative of \( \log x \) is \( 1/x \). Differentiating \( 1/x \) again gives \( -1/x^2 \).

Exam Tip: Remember the standard derivative of \( \log x \) and use the power rule for negative exponents. Always write \( \frac{1}{x} \) as \( x^{-1} \) for easier differentiation.

Question 5. Find the second derivative of \( x^3 \log x \).
Answer: Let \( y = x^3 \log x \).
To find the first derivative, we use the product rule:
\( \frac { dy }{ dx } = x^3 \cdot \frac { d }{ dx } (\log x) + \log x \cdot \frac { d }{ dx } (x^3) \)
\( = x^3 \cdot \frac { 1 }{ x } + \log x \cdot 3x^2 \)
\( = x^2 + 3x^2 \log x \)
Now, we calculate the second derivative by differentiating \( x^2 + 3x^2 \log x \). This involves differentiating each term and using the product rule for the second term:
\( \frac{d^{2} y}{d x^{2}} = \frac { d }{ dx }(x^2) + \frac { d }{ dx } (3x^2 \log x) \)
\( = 2x + [3x^2 \cdot \frac { 1 }{ x } + \log x \cdot 6x] \)
\( = 2x + 3x + 6x \log x \)
\( = 5x + 6x \log x \)
\( = x(5 + 6 \log x) \).
In simple words: We apply the product rule for the first derivative. Then, we differentiate each part of that result, using the product rule again for the term \( 3x^2 \log x \), and then combine similar terms.

Exam Tip: Break down complex functions into smaller, manageable parts. Be careful when applying the product rule multiple times and combining like terms.

Question 6. Find the second derivative of \( e^x \sin 5x \).
Answer: Let \( y = e^x \sin 5x \).
To find the first derivative, we use the product rule:
\( \frac { dy }{ dx } = e^x \frac { d }{ dx }(\sin 5x) + \sin 5x \frac { d }{ dx }e^x \)
\( = e^x (\cos 5x \cdot 5) + \sin 5x \cdot e^x \)
\( = e^x(5 \cos 5x + \sin 5x) \)
Now, we calculate the second derivative by differentiating \( e^x(5 \cos 5x + \sin 5x) \), using the product rule again:
\( \frac{d^{2} y}{d x^{2}} = e^x \frac { d }{ dx }(5 \cos 5x + \sin 5x) + (5 \cos 5x + \sin 5x) \frac { d }{ dx }(e^x) \)
\( = e^x (-5 \sin 5x \cdot 5 + \cos 5x \cdot 5) + (5 \cos 5x + \sin 5x) e^x \)
\( = e^x (-25 \sin 5x + 5 \cos 5x) + e^x (5 \cos 5x + \sin 5x) \)
\( = e^x (-25 \sin 5x + 5 \cos 5x + 5 \cos 5x + \sin 5x) \)
\( = e^x (-24 \sin 5x + 10 \cos 5x) \)
\( = 2e^x (5 \cos 5x - 12 \sin 5x) \).
In simple words: We use the product rule to get the first derivative, remembering the chain rule for \( \sin 5x \). Then, we apply the product rule a second time to differentiate that result, combining terms at the end.

Exam Tip: When using the product rule with exponential and trigonometric functions, remember to apply the chain rule for terms like \( \sin 5x \) or \( \cos 3x \). Factor out \( e^x \) early to simplify calculations.

Question 7. Find the second derivative of \( e^{6x} \cos 3x \).
Answer: Let \( y = e^{6x} \cos 3x \).
First, we find the initial derivative using the product rule and chain rule:
\( \frac { dy }{ dx } = e^{6x} \frac { d }{ dx }(\cos 3x) + \cos 3x \frac { d }{ dx }e^{6x} \)
\( = e^{6x}(- \sin 3x \cdot 3) + \cos 3x (e^{6x} \cdot 6) \)
\( = e^{6x}(-3 \sin 3x + 6 \cos 3x) \)
Now, we calculate the second derivative by differentiating \( e^{6x}(-3 \sin 3x + 6 \cos 3x) \), using the product rule again:
\( \frac{d^{2} y}{d x^{2}} = e^{6x} \frac { d }{ dx }(-3 \sin 3x + 6 \cos 3x) + (-3 \sin 3x + 6 \cos 3x) \frac { d }{ dx }e^{6x} \)
\( = e^{6x}(-3 \cos 3x \cdot 3 - 6 \sin 3x \cdot 3) + (-3 \sin 3x + 6 \cos 3x) e^{6x} \cdot 6 \)
\( = e^{6x}(-9 \cos 3x - 18 \sin 3x) + e^{6x}(-18 \sin 3x + 36 \cos 3x) \)
\( = e^{6x}(-9 \cos 3x - 18 \sin 3x - 18 \sin 3x + 36 \cos 3x) \)
\( = e^{6x}(27 \cos 3x - 36 \sin 3x) \)
\( = 9e^{6x}(3 \cos 3x - 4 \sin 3x) \).
In simple words: We apply the product rule twice, being careful with the chain rule for both \( e^{6x} \) and the trigonometric functions. Remember to group common terms with \( e^{6x} \) at the end.

Exam Tip: Be very meticulous with chain rule applications when differentiating \( e^{ax} \) and \( \sin(bx)/\cos(bx) \). Factoring out \( e^{6x} \) can make the algebraic simplification clearer.

Question 8. Find the second derivative of \( \tan^{-1} x \).
Answer: Let \( y = \tan^{-1} x \).
First, we find the initial derivative of \( \tan^{-1} x \):
\( \frac { dy }{ dx } = \frac { 1 }{ 1+x^2 } = (1+x^2)^{-1} \)
Next, we calculate the second derivative by differentiating \( (1+x^2)^{-1} \) using the chain rule:
\( \frac{d^{2} y}{d x^{2}} = -1 \cdot (1+x^2)^{-2} \cdot \frac { d }{ dx }(1+x^2) \)
\( = -1 \cdot (1+x^2)^{-2} \cdot (2x) \)
\( = \frac{-2x}{(1+x^2)^2} \).
In simple words: We first find the standard derivative of inverse tangent. Then, we treat that result as a power function and use the chain rule to find its derivative again.

Exam Tip: Know the standard derivatives of inverse trigonometric functions. When differentiating \( (f(x))^n \), use the chain rule: \( n(f(x))^{n-1} \cdot f'(x) \).

Question 9. Find the second derivative of \( \log (\log x) \).
Answer: Let \( y = \log (\log x) \).
First, we find the initial derivative using the chain rule:
\( \frac { dy }{ dx } = \frac { 1 }{ \log x } \cdot \frac { d }{ dx }(\log x) \)
\( = \frac { 1 }{ \log x } \cdot \frac { 1 }{ x } \)
\( = (x \log x)^{-1} \)
Next, we calculate the second derivative by differentiating \( (x \log x)^{-1} \) using the chain rule and product rule:
\( \frac{d^{2} y}{d x^{2}} = -1 \cdot (x \log x)^{-2} \cdot \frac { d }{ dx }(x \log x) \)
\( = - \frac { 1 }{ (x \log x)^2 } \cdot \left[x \cdot \frac { d }{ dx }(\log x) + \log x \cdot \frac { d }{ dx }(x)\right] \)
\( = - \frac { 1 }{ (x \log x)^2 } \cdot \left[x \cdot \frac { 1 }{ x } + \log x \cdot 1\right] \)
\( = - \frac { 1 }{ (x \log x)^2 } (1 + \log x) \)
\( = - \frac { 1 + \log x }{ (x \log x)^2 } \).
In simple words: First, we use the chain rule to differentiate \( \log(\log x) \). Then, we use the chain rule again, but this time we also need the product rule for the inner part of the differentiation.

Exam Tip: Nested functions like \( \log (\log x) \) require careful application of the chain rule. Remember to differentiate from the outside in, and use the product rule if necessary for the inner function's derivative.

Question 10. Find the second derivative of \( \sin (\log x) \).
Answer: Let \( y = \sin (\log x) \).
First, we find the initial derivative using the chain rule:
\( \frac { dy }{ dx } = \cos (\log x) \cdot \frac { d }{ dx }(\log x) \)
\( = \cos (\log x) \cdot \frac { 1 }{ x } \)
\( = \frac{\cos(\log x)}{x} \)
Next, we calculate the second derivative by differentiating \( \frac{\cos(\log x)}{x} \) using the quotient rule:
\( \frac{d^{2} y}{d x^{2}} = \frac { x \cdot \frac { d }{ dx }[\cos(\log x)] - \cos(\log x) \cdot \frac { d }{ dx }(x) }{ x^2 } \)
\( = \frac { x \cdot [-\sin(\log x) \cdot \frac { 1 }{ x }] - \cos(\log x) \cdot 1 }{ x^2 } \)
\( = \frac { -\sin(\log x) - \cos(\log x) }{ x^2 } \)
\( = - \frac { \sin(\log x) + \cos(\log x) }{ x^2 } \).
In simple words: First, we differentiate \( \sin(\log x) \) using the chain rule. Then, we differentiate that result using the quotient rule, making sure to apply the chain rule again for \( \cos(\log x) \).

Exam Tip: Be careful with the chain rule for \( \cos(\log x) \) when applying the quotient rule. The derivative of \( \log x \) is \( \frac{1}{x} \), which often simplifies terms in the numerator.

Question 11. If \( y = 5 \cos x - 3 \sin x \), prove that \( \frac{d^{2} y}{d x^{2}} + y = 0 \).
Answer: We have: \( y = 5 \cos x - 3 \sin x \).
First, we find the initial derivative:
\( \frac { dy }{ dx } = -5 \sin x - 3 \cos x \)
Next, we calculate the second derivative:
\( \frac{d^{2} y}{d x^{2}} = -5 \cos x + 3 \sin x \)
We can factor out -1 from the second derivative:
\( = -(5 \cos x - 3 \sin x) \)
From the original equation, we know that \( y = 5 \cos x - 3 \sin x \). So, we can substitute \( y \) into the expression:
\( \frac{d^{2} y}{d x^{2}} = -y \)
Rearranging this equation, we get:
\( \frac{d^{2} y}{d x^{2}} + y = 0 \).
This proves the given statement.
In simple words: We differentiate \( y \) twice. When we get the second derivative, we notice it's just the negative of the original \( y \). Adding them together then results in zero.

Exam Tip: For "prove that" questions, clearly show each differentiation step. Look for opportunities to substitute the original function back into your derivatives to simplify the proof.

Question 12. If \( y = \cos^{-1} x \), find \( \frac{d^{2} y}{d x^{2}} \) in terms of \( y \) alone.
Answer: We have: \( y = \cos^{-1} x \).
This means \( x = \cos y \).
First, we find the initial derivative with respect to \( x \):
\( \frac { dy }{ dx } = \frac { -1 }{ \sqrt{1-x^2} } \)
We can also express this in terms of \( y \). Since \( x = \cos y \), \( \sqrt{1-x^2} = \sqrt{1-\cos^2 y} = \sqrt{\sin^2 y} = \sin y \).
So, \( \frac { dy }{ dx } = \frac { -1 }{ \sin y } = -\operatorname{cosec} y \).
Now, we calculate the second derivative \( \frac{d^{2} y}{d x^{2}} \). We differentiate \( \frac { dy }{ dx } \) with respect to \( x \). It is easier to differentiate \( -\operatorname{cosec} y \) with respect to \( y \) and then multiply by \( \frac{dy}{dx} \):
\( \frac{d^{2} y}{d x^{2}} = \frac { d }{ dy }(-\operatorname{cosec} y) \cdot \frac { dy }{ dx } \)
\( = - (-\operatorname{cosec} y \cot y) \cdot (-\operatorname{cosec} y) \)
\( = - \operatorname{cosec}^2 y \cot y \).
Alternatively, using \( \frac { dy }{ dx } = (1-x^2)^{-1/2} \):
\( \frac{d^{2} y}{d x^{2}} = -\frac{1}{2}(1-x^2)^{-3/2}(-2x) \)
\( = x(1-x^2)^{-3/2} \)
Substitute \( x = \cos y \) and \( \sqrt{1-x^2} = \sin y \):
\( = \frac { \cos y }{ (\sin y)^3 } \)
\( = \frac { \cos y }{ \sin y \cdot \sin^2 y } \)
\( = \cot y \cdot \operatorname{cosec}^2 y \).
The two methods give equivalent results. The negative sign depends on how you handle \( \sqrt{\sin^2 y} = |\sin y| \). Assuming \( y \) is in the principal value range for \( \cos^{-1} x \), \( \sin y \ge 0 \).
Let's recheck the derivative using implicit differentiation from \( x = \cos y \):
\( 1 = -\sin y \frac{dy}{dx} \)
\( \frac{dy}{dx} = -\frac{1}{\sin y} = -\operatorname{cosec} y \)
Differentiating again with respect to \( x \):
\( \frac{d^2y}{dx^2} = \frac{d}{dx}(-\operatorname{cosec} y) \)
\( = \operatorname{cosec} y \cot y \frac{dy}{dx} \)
\( = \operatorname{cosec} y \cot y (-\operatorname{cosec} y) \)
\( = -\operatorname{cosec}^2 y \cot y \).
In simple words: First, we express \( x \) in terms of \( y \). Then, we find the first derivative. To get the second derivative, we differentiate the first derivative with respect to \( x \), remembering to use the chain rule to convert \( dy/dy \) to \( dy/dx \). Finally, we replace \( x \) with \( \cos y \) to have the answer solely in terms of \( y \).

Exam Tip: When asked to express the derivative in terms of \( y \) alone, convert all \( x \) terms using the original relationship \( x = \cos y \). Implicit differentiation can often simplify the process.

Question 13. If \( y = 3 \cos (\log x) + 4 \sin (\log x) \), show that \( x^2 y_2 + xy_1 + y = 0 \).
Answer: We have: \( y = 3 \cos (\log x) + 4 \sin (\log x) \quad ...(1) \)
First, we find \( y_1 = \frac{dy}{dx} \) using the chain rule:
\( y_1 = 3[-\sin (\log x) \cdot \frac{1}{x}] + 4[\cos (\log x) \cdot \frac{1}{x}] \)
\( y_1 = -\frac{3 \sin (\log x)}{x} + \frac{4 \cos (\log x)}{x} \)
Multiply by \( x \) to remove the fraction:
\( xy_1 = -3 \sin (\log x) + 4 \cos (\log x) \)
Next, we differentiate \( xy_1 \) again with respect to \( x \). Use the product rule for \( xy_1 \):
\( x \frac{d}{dx}(y_1) + y_1 \frac{d}{dx}(x) = -3[\cos (\log x) \cdot \frac{1}{x}] + 4[-\sin (\log x) \cdot \frac{1}{x}] \)
\( xy_2 + y_1 \cdot 1 = -\frac{3 \cos (\log x)}{x} - \frac{4 \sin (\log x)}{x} \)
\( xy_2 + y_1 = -\frac{1}{x}[3 \cos (\log x) + 4 \sin (\log x)] \)
From (1), we know that \( y = 3 \cos (\log x) + 4 \sin (\log x) \). Substitute \( y \) into the equation:
\( xy_2 + y_1 = -\frac{y}{x} \)
Multiply the entire equation by \( x \) to clear the fraction:
\( x^2 y_2 + xy_1 = -y \)
Rearrange the terms to get the desired form:
\( x^2 y_2 + xy_1 + y = 0 \).
This proves the statement.
In simple words: We find the first derivative and multiply by \( x \). Then, we find the second derivative using the product rule. We notice that the expression appearing in the second derivative is the original \( y \). Substituting \( y \) and rearranging the terms gives us the required proof.

Exam Tip: When proving identities with higher-order derivatives, simplify expressions after each differentiation step, especially by multiplying by \( x \) if \( 1/x \) terms appear. Look for opportunities to substitute the original function back into the derivatives.

Question 14. If \( y = Ae^{mx} + Be^{nx} \), show that \( \frac{d^{2} y}{d x^{2}} - (m + n)\frac { dy }{ dx } + mny = 0 \).
Answer: We have: \( y = Ae^{mx} + Be^{nx} \).
First, we find the initial derivative \( \frac{dy}{dx} \):
\( \frac { dy }{ dx } = A \cdot me^{mx} + B \cdot ne^{nx} \)
Next, we calculate the second derivative \( \frac{d^{2} y}{d x^{2}} \):
\( \frac{d^{2} y}{d x^{2}} = A \cdot m^2 e^{mx} + B \cdot n^2 e^{nx} \)
Now, we substitute these derivatives into the given equation:
\( \frac{d^{2} y}{d x^{2}} - (m + n)\frac { dy }{ dx } + mny \)
\( = (Am^2 e^{mx} + Bn^2 e^{nx}) - (m + n)(Ame^{mx} + Bne^{nx}) + mn(Ae^{mx} + Be^{nx}) \)
Expand the terms:
\( = Am^2 e^{mx} + Bn^2 e^{nx} - (m \cdot Ame^{mx} + m \cdot Bne^{nx} + n \cdot Ame^{mx} + n \cdot Bne^{nx}) + mnAe^{mx} + mnBe^{nx} \)
\( = Am^2 e^{mx} + Bn^2 e^{nx} - Am^2 e^{mx} - mnBe^{nx} - mnAe^{mx} - Bn^2 e^{nx} + mnAe^{mx} + mnBe^{nx} \)
Now, we group and cancel out similar terms:
\( = (Am^2 e^{mx} - Am^2 e^{mx}) + (Bn^2 e^{nx} - Bn^2 e^{nx}) + (-mnBe^{nx} + mnBe^{nx}) + (-mnAe^{mx} + mnAe^{mx}) \)
\( = 0 + 0 + 0 + 0 \)
\( = 0 \).
Thus, \( \frac{d^{2} y}{d x^{2}} - (m + n)\frac { dy }{ dx } + mny = 0 \).
In simple words: We find the first and second derivatives of the given \( y \). Then, we put these into the equation we need to prove. After expanding and grouping, all the terms cancel out, showing the expression equals zero.

Exam Tip: Be very careful with algebraic expansion and cancellation of terms. It's helpful to write out each term clearly to avoid sign errors or missing terms. Each derivative uses the chain rule in the form \( \frac{d}{dx}(e^{kx}) = ke^{kx} \).

Question 15. If \( y = 500 e^{7x} + 600 e^{-7x} \), show that \( \frac{d^{2} y}{d x^{2}} = 49y \).
Answer: We have: \( y = 500 e^{7x} + 600 e^{-7x} \quad ...(1) \)
First, we find the initial derivative \( \frac{dy}{dx} \):
\( \frac { dy }{ dx } = 500 \cdot 7e^{7x} + 600 \cdot (-7)e^{-7x} \)
\( = 3500e^{7x} - 4200e^{-7x} \)
Next, we calculate the second derivative \( \frac{d^{2} y}{d x^{2}} \):
\( \frac{d^{2} y}{d x^{2}} = 3500 \cdot 7e^{7x} - 4200 \cdot (-7)e^{-7x} \)
\( = 24500e^{7x} + 29400e^{-7x} \)
Now, we can factor out 49 from this expression:
\( = 49 (500e^{7x} + 600e^{-7x}) \)
From (1), we know that \( y = 500e^{7x} + 600e^{-7x} \). Substitute \( y \) into the expression:
\( \frac{d^{2} y}{d x^{2}} = 49y \).
This proves the given statement.
In simple words: We differentiate \( y \) twice, remembering the chain rule for \( e^{7x} \) and \( e^{-7x} \). After getting the second derivative, we factor out 49 and see that the remaining expression is exactly the original \( y \).

Exam Tip: For problems involving exponential functions like \( e^{ax} \), the derivative is \( ae^{ax} \). Notice the pattern in the coefficients after each differentiation, which often leads to a multiple of the original function.

Question 16. If \( e^y (x+1) = 1 \), show that \( \frac{d^{2} y}{d x^{2}} = (\frac { dy }{ dx })^2 \).
Answer: We have: \( e^y (x+1) = 1 \).
First, we can write \( e^y = \frac{1}{x+1} = (x+1)^{-1} \).
Differentiate both sides with respect to \( x \):
\( \frac{d}{dx}(e^y) (x+1) + e^y \frac{d}{dx}(x+1) = \frac{d}{dx}(1) \)
\( e^y \frac{dy}{dx} (x+1) + e^y (1) = 0 \)
Divide by \( e^y \) (since \( e^y \neq 0 \)):
\( (x+1)\frac{dy}{dx} + 1 = 0 \)
\( (x+1)\frac{dy}{dx} = -1 \)
\( \frac{dy}{dx} = -\frac{1}{x+1} \)
From \( e^y = \frac{1}{x+1} \), we can substitute this back:
\( \frac{dy}{dx} = -e^y \quad ...(1) \)
Now, we differentiate \( \frac{dy}{dx} = -e^y \) again with respect to \( x \) to find \( \frac{d^{2} y}{d x^{2}} \):
\( \frac{d^{2} y}{d x^{2}} = \frac{d}{dx}(-e^y) \)
\( = -e^y \frac{dy}{dx} \)
From (1), we know that \( \frac{dy}{dx} = -e^y \). So, \( -e^y = \frac{dy}{dx} \).
Substitute this into the expression for \( \frac{d^{2} y}{d x^{2}} \):
\( \frac{d^{2} y}{d x^{2}} = \left(\frac{dy}{dx}\right) \frac{dy}{dx} \)
\( \frac{d^{2} y}{d x^{2}} = \left(\frac{dy}{dx}\right)^2 \).
This proves the statement.
In simple words: We first find the first derivative using implicit differentiation. Then, we differentiate that result again. By carefully substituting back the first derivative, we show that the second derivative is the square of the first derivative.

Exam Tip: For implicit differentiation, remember to apply the chain rule when differentiating terms involving \( y \) with respect to \( x \). Simplifying the first derivative can often make calculating the second derivative much easier.

Question 17. If \( y = (\tan^{-1} x)^2 \), show that \( (x^2+1)^2 y_2 + 2x(x^2+1) y_1 = 2 \).
Answer: We have: \( y = (\tan^{-1} x)^2 \).
First, we find \( y_1 = \frac{dy}{dx} \) using the chain rule:
\( y_1 = 2(\tan^{-1} x) \cdot \frac{d}{dx}(\tan^{-1} x) \)
\( y_1 = 2(\tan^{-1} x) \cdot \frac{1}{1+x^2} \)
Multiply by \( (1+x^2) \) to clear the fraction:
\( (1+x^2)y_1 = 2 \tan^{-1} x \)
Next, we differentiate \( (1+x^2)y_1 = 2 \tan^{-1} x \) again with respect to \( x \). Use the product rule for the left side:
\( (1+x^2) \frac{d}{dx}(y_1) + y_1 \frac{d}{dx}(1+x^2) = 2 \frac{d}{dx}(\tan^{-1} x) \)
\( (1+x^2)y_2 + y_1(2x) = 2 \cdot \frac{1}{1+x^2} \)
Now, multiply the entire equation by \( (1+x^2) \) to clear the remaining fraction:
\( (1+x^2)^2 y_2 + 2x(1+x^2)y_1 = 2 \).
This proves the given statement.
In simple words: We find the first derivative of \( y \) using the chain rule. Then, we simplify this equation by removing the fraction. Next, we differentiate this new equation again, using the product rule. Finally, we multiply by \( (1+x^2) \) to eliminate fractions and arrive at the desired result.

Exam Tip: When dealing with inverse trigonometric functions squared, remember the chain rule for the outer power and then the derivative of the inverse function. Multiply by denominators early to simplify subsequent differentiation steps.