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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Differentiation

NDA 2019 · Mathematics · Differentiation

What is the derivative of 2^ ( x)^2 with respect to x ?

What is the derivative of 2(sin⁡x)22^{(\sin x)^2} with respect to sin⁡x\sin x ?

Options

  1. Asin⁡x2(sin⁡x)2ln⁡4\sin x 2^{(\sin x)^2} \ln 4
  2. B2sin⁡x2(sin⁡x)2ln⁡42 \sin x 2^{(\sin x)^2} \ln 4
  3. CIn⁡(sin⁡x)2(sin⁡x)2\operatorname{In}(\sin x) 2^{(\sin x)^2}
  4. D2sin⁡xcos⁡x2(sin⁡x)22 \sin x \cos x 2^{(\sin x)^2}

Correct answer

B. 2sin⁡x2(sin⁡x)2ln⁡42 \sin x 2^{(\sin x)^2} \ln 4

Step-by-step solution

y=2sin⁡2x⇒ln⁡y=sin⁡2x⋅ln⁡2⇒1y⋅dydx=2sin⁡x⋅cos⁡x⋅ln⁡2⇒dydx=y⋅sin⁡x⋅cos⁡x⋅ln⁡4\begin{aligned} & y=2^{\sin ^2 x} \\ & \Rightarrow \ln y=\sin ^2 x \cdot \ln 2 \\ & \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=2 \sin x \cdot \cos x \cdot \ln 2 \\ & \Rightarrow \frac{d y}{d x}=y \cdot \sin x \cdot \cos x \cdot \ln 4 \end{aligned} Let sin⁡x=4\sin x=4∴dydx=cos⁡x∴dydx=ℓn4⋅sin⁡x⋅2sin⁡2x⋅cos⁡xcos⁡x⇒dydx=2sin⁡2x⋅sin⁡x⋅ℓn4\begin{aligned} & \therefore \quad \frac{d y}{d x}=\cos x \\ & \therefore \quad \frac{d y}{d x}=\frac{\ell \mathrm{n} 4 \cdot \sin x \cdot 2^{\sin ^2 x} \cdot \cos x}{\cos x} \\ & \Rightarrow \frac{d y}{d x}=2^{\sin ^2 x} \cdot \sin x \cdot \ell \mathrm{n} 4 \end{aligned}$$
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