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Question: Differentiate \({{x}^{\sin x}},x>0\) with respect to \(x\) ....

Differentiate xsin⁡x,x>0{{x}^{\sin x}},x>0 with respect to xx .

Explanation

Solution

Start by using y=xsin⁡xy={{x}^{\sin x}} and taking log of both the sides of the equation. Now use the identity that log⁡ab=blog⁡a\log {{a}^{b}}=b\log a and differentiate both sides of the equation with respect to x. Use the uv rule of differentiation, i.e., d(uv)dx=vdudx+udvdx\dfrac{d\left( uv \right)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx} .

Complete step-by-step solution -
Let us start the solution to the above question by letting xsin⁡x,x>0{{x}^{\sin x}},x>0 to be equal to y. So, we can say
y=xsin⁡x............(i)y={{x}^{\sin x}}............(i)
Now we know, if we take log of both sides of the equation, we know that the equation remains valid. So, we will take log of both sides of the equation. On doing so, we get
log⁡y=log⁡xsin⁡x\log y=\log {{x}^{\sin x}}
Now, we know that log⁡ab=blog⁡a\log {{a}^{b}}=b\log a . So, using this identity in our equation, we get
log⁡y=sin⁡xlog⁡x\log y=\sin x\log x
Now we will differentiate both sides of the equation with respect to x. On doing so, we get
d(log⁡y)dx=d(sin⁡xlog⁡x)dx\dfrac{d\left( \log y \right)}{dx}=\dfrac{d\left( \sin x\log x \right)}{dx}
Using the uv rule of differentiation, i.e., d(uv)dx=vdudx+udvdx\dfrac{d\left( uv \right)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx} and d(log⁡y)dx=1ydydx\dfrac{d\left( \log y \right)}{dx}=\dfrac{1}{y}\dfrac{dy}{dx} , we get
1ydydx=log⁡xd(sin⁡x)dx+sin⁡xd(log⁡x)dx\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{\log xd\left( \sin x \right)}{dx}+\dfrac{\sin xd\left( \log x \right)}{dx}
Now, we know that the derivative of sin⁡x is cosx\sin x\text{ is cosx} and log⁡x is 1x\log x\text{ is }\dfrac{1}{x} . So, using this in our equation, we get
1ydydx=log⁡x×cosx⁡+sin⁡x×1x\dfrac{1}{y}\dfrac{dy}{dx}=\log x\times \operatorname{cosx}+\sin x\times \dfrac{1}{x}
⇒dydx=y(log⁡x×cosx⁡+sin⁡x×1x)\Rightarrow \dfrac{dy}{dx}=y\left( \log x\times \operatorname{cosx}+\sin x\times \dfrac{1}{x} \right)
Now, we will put the value of y from equation (i). On doing so, we get
dydx=xsin⁡x(cosxlog⁡x+sin⁡xx)\dfrac{dy}{dx}={{x}^{\sin x}}\left( cosx\log x+\dfrac{\sin x}{x} \right)
Therefore, we can conclude that the derivative of xsin⁡x,x>0{{x}^{\sin x}},x>0 is xsin⁡x(cosxlog⁡x+sin⁡xx).{{x}^{\sin x}}\left( cosx\log x+\dfrac{\sin x}{x} \right).

Note: Be careful with the signs and calculations as in such questions, the possibility of making a mistake is either of the sign or a calculation error. Also, remember that for taking log of both the sides of the equation, both the sides must be positive. In the above question the RHS is positive as it is given that x>0, if it was not mentioned, we could not have taken log of both the sides and solved the equation.