Question

How do you differentiate ${{x}^{\sin \left( x \right)}}$?

Answer 1

In this problem we have to find the differentiation value of ${{x}^{\sin \left( x \right)}}$. We assume the given expression as $y={{x}^{\sin \left( x \right)}}$. Now we will take natural logarithms on both sides of the above equation. We will simplify the logarithmic expression by using the logarithmic formulas. Now we will differentiate to both sides of the equation with respect to the $''x''$. Now we will implicit differentiation on the LHS. We will use the product rule in RHS. The product rule is$\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}\left( v \right)+v\dfrac{d}{dx}\left( u \right)$. Now we will simplify the expression then we will get the result.

Formulas Used:
1.$\log {{\left( x \right)}^{y}}=y\log x$
2.$\dfrac{d}{dx}(\log y)=\dfrac{1}{y}\dfrac{dy}{dx}$
3.$\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$
4.We use product rule of differentiation
$\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}\left( v \right)+v\dfrac{d}{dx}\left( u \right)$
5.$\dfrac{d}{dx}\left( \sin x \right)=\cos x$.

Complete step by step solution:
Given that, ${{x}^{\sin x}}$
Now we will write the above expression $y$ is equal to ${{x}^{\sin x}}$,then
$y=$ ${{x}^{\sin x}}$
Now we will take natural logarithms to both sides, then,
$\log y=\log {{\left( x \right)}^{\sin x}}$
Using the formula $\log {{\left( x \right)}^{y}}=y\log x$ in the above equation, then we will get
$\log y=\sin x\log x$
Now differentiate the above equation with respect $x$
$\dfrac{d}{dx}\left( \log y \right)=\dfrac{d}{dx}\left( \sin x\log x \right)$
Using the formula $\dfrac{d}{dx}(\log y)=\dfrac{1}{y}\dfrac{dy}{dx}$ in the above equation, then we will get
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{dy}{dx}\left( \sin x\log x \right)$
Using the product rule of differentiation $\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}\left( v \right)+v\dfrac{d}{dx}\left( u \right)$ in the above equation, then we will get
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\log x\dfrac{d}{dx}\sin x+\sin x\dfrac{d}{dx}\log x$
We know that $\dfrac{d}{dx}\left( \sin x \right)=\cos x$,then
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\cos x\log x+\sin x\dfrac{d}{dx}\log x$
We know that $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$,then
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\cos x\log x+\sin x\dfrac{1}{x}$
Now we will simplify the above equation, then
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\cos x\log x+\dfrac{\sin x}{x}$
$\Rightarrow \dfrac{dy}{dx}=y\left( \cos x\log x+\dfrac{\sin x}{x} \right)$
We know that $y=$ ${{x}^{\sin x}}$ then substitute in the above equation, then we will get
$\Rightarrow \dfrac{dy}{dx}={{x}^{\sin x}}\left( \cos x\log x+\dfrac{\sin x}{x} \right)$
Hence the differentiation value if

$\therefore \dfrac{dy}{dx}=\left( {{x}^{\sin x}} \right)\left( \cos x\log x+\dfrac{\sin x}{x} \right)$

Note:
This is similar to the product rule of differentiation. Basically, in product rule what we do is, we treat $u$constant and differentiation $v$ and then take $v$ constant and differentiate $u$and add them together. similarly if we take this function ,it is of the form ${{u}^{v}}$.so when we take $u$as constant .it become a differentiation of form ${{a}^{x}}$(which will give ${{a}^{x}}\log a$) (also don’t forget to differentiate $\sin x$) and then when we take $v$ as a constant, it will be of the form ${{x}^{n}}$(which will give $nx\left( n-1 \right)$).