Maharashtra Board Class 12 Maths Part 1 Chapter 3 Differentiation 3.6 Solutions

1. Find \( \frac{d^2y}{dx^2} \) if,

Question 1. \( y = \sqrt{x} \)
Answer:
Solution:
\( y = \sqrt{x} \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}} \)
Differentiating again w.r.t. x, we get
\( \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1}{2\sqrt{x}} \right) = \frac{1}{2} \frac{d}{dx} (x^{-\frac{1}{2}}) \)
\( \implies = \frac{1}{2} \left( -\frac{1}{2} \right) x^{-\frac{1}{2}-1} = -\frac{1}{4} x^{-\frac{3}{2}} \)
In simple words: To find the second derivative of \( \sqrt{x} \), first differentiate it once to get \( \frac{1}{2\sqrt{x}} \), then differentiate this result again. This involves using the power rule for derivatives after rewriting \( \sqrt{x} \) as \( x^{1/2} \) and \( \frac{1}{2\sqrt{x}} \) as \( \frac{1}{2}x^{-1/2} \).

🎯 Exam Tip: Remember to express square roots as fractional exponents before differentiating, and be careful with negative exponents during the second differentiation step to avoid common errors.

Question 2. \( y = x^5 \)
Answer:
Solution:
\( y = x^5 \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \frac{d}{dx} (x^5) = 5x^4 \)
Differentiating again w.r.t. x, we get
\( \frac{d^2y}{dx^2} = \frac{d}{dx} (5x^4) = 5\frac{d}{dx} (x^4) \)
\( \implies = 5 \times 4x^3 = 20x^3 \)
In simple words: To find the second derivative of \( x^5 \), apply the power rule for differentiation twice. The first derivative is \( 5x^4 \), and differentiating this expression again yields \( 20x^3 \).

🎯 Exam Tip: The power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) is fundamental for these types of problems. Ensure you apply it correctly for successive differentiations.

Question 3. \( y = x^{-7} \)
Answer:
Solution:
\( y = x^{-7} \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \frac{d}{dx} (x^{-7}) = -7x^{-8} \)
Differentiating again w.r.t. x, we get
\( \frac{d^2y}{dx^2} = \frac{d}{dx} (-7x^{-8}) = -7\frac{d}{dx} (x^{-8}) \)
\( \implies = (-7)(-8)x^{-9} = 56x^{-9} \)
In simple words: For \( y = x^{-7} \), the first derivative is found by bringing the power down and subtracting one from the exponent, resulting in \( -7x^{-8} \). Repeating this process for the first derivative gives the second derivative, which is \( 56x^{-9} \).

🎯 Exam Tip: Pay close attention to the signs when dealing with negative exponents. A common mistake is an error in multiplying negative numbers or subtracting 1 from a negative exponent.

2. Find \( \frac{d^2y}{dx^2} \) if,

Question 1. \( y = e^x \)
Answer:
Solution:
\( y = e^x \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \frac{d}{dx} (e^x) = e^x \)
Differentiating again w.r.t. x, we get
\( \frac{d^2y}{dx^2} = \frac{d}{dx} (e^x) = e^x \)
In simple words: The derivative of \( e^x \) with respect to \( x \) is \( e^x \). Therefore, its second derivative is also \( e^x \), demonstrating a unique property of the exponential function.

🎯 Exam Tip: Remember that \( \frac{d}{dx}(e^x) = e^x \) is a fundamental derivative formula. This property simplifies finding higher-order derivatives for this function considerably.

Question 2. \( y = e^{(2x+1)} \)
Answer:
Solution:
\( y = e^{(2x+1)} \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \frac{d}{dx} [e^{(2x+1)}] = e^{(2x+1)} \cdot \frac{d}{dx} (2x+1) \)
\( \implies = e^{(2x+1)} \times (2 \times 1 + 0) = 2e^{(2x+1)} \)
Differentiating again w.r.t. x, we get
\( \frac{d^2y}{dx^2} = \frac{d}{dx} [2e^{(2x+1)}] = 2\frac{d}{dx} [e^{(2x+1)}] \)
\( \implies = 2e^{(2x+1)} \cdot \frac{d}{dx} (2x+1) = 2e^{(2x+1)} \times (2 \times 1 + 0) \)
\( \implies = 4e^{(2x+1)} \)
In simple words: To find the derivatives of \( e^{(2x+1)} \), use the chain rule. The first derivative is \( 2e^{(2x+1)} \) (derivative of \( e^u \) is \( e^u \cdot u' \)). Apply the chain rule again to this result to get the second derivative, which is \( 4e^{(2x+1)} \).

🎯 Exam Tip: For composite exponential functions like \( e^{f(x)} \), the chain rule is crucial. Ensure you correctly multiply by the derivative of the exponent in each differentiation step.

Question 3. \( y = e^{\log x} \)
Answer:
Solution:
\( y = e^{\log x} \)
\[ \because a^{\log_a x} = x \] So, \( y = x \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \frac{d}{dx} (x) = 1 \)
Differentiating again w.r.t. x, we get
\( \frac{d^2y}{dx^2} = \frac{d}{dx} (1) = 0 \)
In simple words: The expression \( e^{\log x} \) simplifies to \( x \) due to the inverse property of exponential and natural logarithm functions. Therefore, its first derivative is 1, and its second derivative is 0.

🎯 Exam Tip: Always simplify expressions using logarithmic and exponential properties before differentiating. This can significantly reduce complexity and potential errors.