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Worked Examples · Example 29

Q.Differentiate xsin⁡xx^{\sin x}, x>0x > 0 w.r.t. xx.

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We differentiate xsin⁡xx^{\sin x} by rewriting it as esin⁡x⋅log⁡xe^{\sin x \cdot \log x} and then applying the chain rule and product rule. The derivative is xsin⁡x(cos⁡x⋅log⁡x+sin⁡xx)x^{\sin x} \left( \cos x \cdot \log x + \frac{\sin x}{x} \right).

The function xsin⁡xx^{\sin x} is not a simple power function (where the exponent is constant) nor a simple exponential (where the base is constant). The variable appears in both the base and the exponent. To handle this, we use a technique called logarithmic differentiation — which is really just implicit differentiation in disguise.

The core idea: take the natural logarithm of both sides, use log properties to bring the exponent down, then differentiate implicitly. This converts the problem into a product rule inside a chain rule, which is much easier to manage.


  1. Set up the equation Let y=xsin⁡xy = x^{\sin x}, with x>0x > 0 (so log⁡x\log x is defined). Take the natural log of both sides:

log⁡y=log⁡(xsin⁡x)=sin⁡x⋅log⁡x\log y = \log\left(x^{\sin x}\right) = \sin x \cdot \log x

  1. Differentiate implicitly with respect to xx On the left, ddx(log⁡y)=1y⋅dydx\frac{d}{dx}(\log y) = \frac{1}{y} \cdot \frac{dy}{dx} (by the chain rule). On the right, sin⁡x⋅log⁡x\sin x \cdot \log x is a product, so we use the product rule:

ddx(sin⁡x⋅log⁡x)=cos⁡x⋅log⁡x+sin⁡x⋅1x\frac{d}{dx}(\sin x \cdot \log x) = \cos x \cdot \log x + \sin x \cdot \frac{1}{x}

Putting it together:

1y⋅dydx=cos⁡x⋅log⁡x+sin⁡xx\frac{1}{y} \cdot \frac{dy}{dx} = \cos x \cdot \log x + \frac{\sin x}{x}

  1. Solve for dydx\frac{dy}{dx} Multiply both sides by yy:

dydx=y(cos⁡x⋅log⁡x+sin⁡xx)\frac{dy}{dx} = y \left( \cos x \cdot \log x + \frac{\sin x}{x} \right)

Now substitute back y=xsin⁡xy = x^{\sin x}:

dydx=xsin⁡x(cos⁡x⋅log⁡x+sin⁡xx)\frac{dy}{dx} = x^{\sin x} \left( \cos x \cdot \log x + \frac{\sin x}{x} \right) …

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