Question

Let y=(tanx)^sinx for \(0\lt x\lt\frac{\pi}{2}\). If \(\frac{dy}{dx}=(\tan x)^{\sin x}((\cos x)\log(\tan x)+g(x))\), then g(x)=

• A. secx
• B. secx cosecx
• C. sinx sec²x
• D. cosecx
• E. sinx tanx

Answer 1

The Correct Option is C

We are given that \( y = (\tan x)^{\sin x} \) for \( 0 < x < \frac{\pi}{2} \). We need to find the function \( g(x) \) such that the derivative \( \frac{dy}{dx} \) is given by:

 \(\frac{dy}{dx} = (\tan x)^{\sin x} \left( (\cos x) \log (\tan x) + g(x) \right)\)

We will begin by differentiating \( y = (\tan x)^{\sin x} \) using the chain rule and logarithmic differentiation.

Let \( y = (\tan x)^{\sin x} \), and take the natural logarithm on both sides:

\(\ln y = \sin x \cdot \ln (\tan x)\)

Now, differentiate both sides with respect to \( x \):

\(\frac{d}{dx}(\ln y) = \frac{d}{dx}(\sin x \cdot \ln (\tan x))\)

On the left-hand side, we use the chain rule:

\(\frac{1}{y} \cdot \frac{dy}{dx}\)

Now, differentiate the right-hand side using the product rule:

\(\frac{d}{dx}(\sin x \cdot \ln (\tan x)) = \cos x \cdot \ln (\tan x) + \sin x \cdot \frac{d}{dx}(\ln (\tan x))\)

Next, differentiate \( \ln (\tan x) \) with respect to \( x \):

\(\frac{d}{dx}(\ln (\tan x)) = \frac{1}{\tan x} \cdot \sec^2 x\)

Thus, the derivative becomes:

\(\frac{dy}{dx} = (\tan x)^{\sin x} \left( (\cos x) \log (\tan x) + \sin x \cdot \sec^2 x \right)\)

Comparing this with the given derivative, we can see that:

\(g(x) = \sin x \cdot \sec^2 x\)

Therefore, the correct answer is:

sin x sec² x