Derivative of x^sinx: Formula, Proof | x^sinx Derivative

The derivative of x^sinx (x to the power sinx) is equal to x^sinx[sinx/x + cosx logx]. Here we learn how to differentiate x^sinx.

Derivative of x^sinx Formula

x^sinx Derivative Formula: The formula of x^sinx derivative is given by

d/dx(x^sinx) = x^sinx[sinx/x + cosx logx]

Let us now give a proof of this fact.

Derivative of x^sinx Proof

Question: Prove that d/dx(x^sinx) = x^sinx[sinx/x + cosx logx].

Answer:

Let y=x^sinx. We need to find dy/dx.

Taking the natural logarithms both sides, we have

log_e y = log_e x^sinx

⇒ log_ey =sinx log_ex.

Taking d/dx on both sides, we have

$\dfrac{d}{dx}(\log_e y)=\dfrac{d}{dx}(\sin x \log_e x)$

Now, applying the product rule of derivatives, it follows that

$\dfrac{1}{y} \dfrac{dy}{dx}$ $=\sin x\dfrac{d}{dx}{\log_e x}+\log_e x\dfrac{d}{dx}(\sin x)$

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ $=\sin x \cdot \dfrac{1}{x}+\log_e x \cos x$ as the derivative of log_ex is 1/x and d/dx(sinx)=cosx.

⇒ $\dfrac{dy}{dx}=y(\dfrac{\sin x}{x}+\cos x\log_e x)$

⇒ $\dfrac{dy}{dx}=x^{\sin x}(\dfrac{\sin x}{x}+\cos x \log_e x)$ as y=x^sinx.

Thus, the derivative of x^sinx (x to the sinx) is equal to x^sinx[sinx/x + cosx logx]. This is proved by the logarithmic differentiation.

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