Question:

If y=(tanx)xy = (\tan x)^x, then find dydx\frac{dy}{dx}.

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For functions of the form y=[f(x)]g(x)y = [f(x)]^{g(x)}, take the natural logarithm on both sides to handle the variable exponent, and then differentiate using the product rule and chain rule.
Updated On: Jan 18, 2025
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Solution and Explanation

The given function is: y=(tanx)x. y = (\tan x)^x.  
Take the natural logarithm on both sides to simplify the power: lny=xln(tanx). \ln y = x \ln (\tan x).  

Differentiate both sides with respect to xx: 1ydydx=ln(tanx)+x1tanxsec2x. \frac{1}{y} \cdot \frac{dy}{dx} = \ln (\tan x) + x \cdot \frac{1}{\tan x} \cdot \sec^2 x.  

Simplify: 1ydydx=ln(tanx)+xsec2xtanx. \frac{1}{y} \cdot \frac{dy}{dx} = \ln (\tan x) + x \cdot \frac{\sec^2 x}{\tan x}.  

Multiply through by y=(tanx)xy = (\tan x)^x: dydx=(tanx)x[ln(tanx)+xsec2xtanx]. \frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \frac{\sec^2 x}{\tan x}\right].  

Simplify further: dydx=(tanx)x[ln(tanx)+x1sinxcosx]. \frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \frac{1}{\sin x \cos x}\right].  

Final answer: dydx=(tanx)x[ln(tanx)+xcscxsecx]. \frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \csc x \sec x\right].

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