▶️ Answer/Explanation
Solution
Correct Answer: A
Step 1: Find the second derivative of y with respect to x:
\[ \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} = \frac{\mathrm{d}}{\mathrm{d} x}\left(x^{4}-2x^{3}+3x-1\right) \] Step 2: Differentiate term by term:
\[ \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} = 4x^{3} – 6x^{2} + 3 \] Step 3: Evaluate at x = 2:
\[ \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}\bigg|_{x=2} = 4(2)^{3} – 6(2)^{2} + 3 = 32 – 24 + 3 = 11 \]
▶️ Answer/Explanation
Solution
Correct Answer: D
Step 1: Find the first derivative using the chain rule:
\[ \frac{dy}{dx} = e^{x^{3}} \cdot \frac{d}{dx}(x^{3}) = 3x^{2}e^{x^{3}} \] Step 2: Find the second derivative using the product rule and chain rule:
\[ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(3x^{2}e^{x^{3}}) \] \[ = 6xe^{x^{3}} + 3x^{2} \cdot 3x^{2}e^{x^{3}} \] \[ = 6xe^{x^{3}} + 9x^{4}e^{x^{3}} \] \[ = (6x + 9x^{4})e^{x^{3}} \]
▶️ Answer/Explanation
Solution
Correct Answer: D
Step 1: Find the first derivative using the chain rule:
\[ \frac{dy}{dx} = e^{2\sin x} \cdot \frac{d}{dx}(2\sin x) = 2\cos x \cdot e^{2\sin x} \] Step 2: Find the second derivative using the product rule:
\[ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(2\cos x \cdot e^{2\sin x}) \] \[ = -2\sin x \cdot e^{2\sin x} + 2\cos x \cdot 2\cos x \cdot e^{2\sin x} \] \[ = (-2\sin x + 4\cos^{2}x)e^{2\sin x} \] \[ = 2(-\sin x + 2\cos^{2}x)e^{2\sin x} \]
Question
If \(y=-3\cos (2x)\) , then \(\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}\)
A −12cos(2x)
B 12cos(2x)
C −3cos(2x)
D 3cos(2x)
▶️Answer/Explanation
Ans:B
The chain rule must be used twice to find the second derivative
\(\frac{\mathrm{d} Y}{\mathrm{d} x}=-3(-\sin (2x)).2\)
\(\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=\frac{\mathrm{d} }{\mathrm{d} x}(6\sin (2x)).2=12\cos 2x\)