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What is the derivative of 2^((sinx)^(2))...

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

A

`sinx.2^((sinx)^(2))ln4`

B

`2sinx.2^((sinx)^(2))ln4`

C

`ln(sinx)2^((sinx)^(2))`

D

`2 sinx cosx2^((sinx)^(2))`

Text Solution

AI Generated Solution

The correct Answer is:
To find the derivative of \( 2^{(\sin x)^2} \) with respect to \( \sin x \), we can follow these steps: ### Step 1: Define the function Let \( u = 2^{(\sin x)^2} \). We want to find \( \frac{du}{d(\sin x)} \). ### Step 2: Use the chain rule By the chain rule, we can express the derivative as: \[ \frac{du}{d(\sin x)} = \frac{du}{dx} \cdot \frac{dx}{d(\sin x)} \] Since \( \frac{dx}{d(\sin x)} = \cos x \), we can rewrite the equation as: \[ \frac{du}{d(\sin x)} = \frac{du}{dx} \cdot \cos x \] ### Step 3: Find \( \frac{du}{dx} \) To find \( \frac{du}{dx} \), we will use the formula for the derivative of an exponential function: \[ \frac{d}{dx}(a^f) = a^f \ln(a) \cdot \frac{df}{dx} \] In our case, \( a = 2 \) and \( f = (\sin x)^2 \). Therefore, \[ \frac{du}{dx} = 2^{(\sin x)^2} \cdot \ln(2) \cdot \frac{d}{dx}((\sin x)^2) \] ### Step 4: Find \( \frac{d}{dx}((\sin x)^2) \) Using the chain rule again, we have: \[ \frac{d}{dx}((\sin x)^2) = 2 \sin x \cdot \cos x \] ### Step 5: Substitute back into \( \frac{du}{dx} \) Now substituting \( \frac{d}{dx}((\sin x)^2) \) into our expression for \( \frac{du}{dx} \): \[ \frac{du}{dx} = 2^{(\sin x)^2} \cdot \ln(2) \cdot (2 \sin x \cos x) \] This simplifies to: \[ \frac{du}{dx} = 2^{(\sin x)^2} \cdot 2 \sin x \cos x \cdot \ln(2) \] ### Step 6: Substitute \( \frac{du}{dx} \) back into the chain rule expression Now we can substitute \( \frac{du}{dx} \) back into our expression for \( \frac{du}{d(\sin x)} \): \[ \frac{du}{d(\sin x)} = \left(2^{(\sin x)^2} \cdot 2 \sin x \cos x \cdot \ln(2)\right) \cdot \cos x \] This simplifies to: \[ \frac{du}{d(\sin x)} = 2^{(\sin x)^2} \cdot 2 \sin x \cos^2 x \cdot \ln(2) \] ### Final Answer Thus, the derivative of \( 2^{(\sin x)^2} \) with respect to \( \sin x \) is: \[ \frac{du}{d(\sin x)} = 2^{(\sin x)^2} \cdot 2 \sin x \cos^2 x \cdot \ln(2) \]
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