Question

If we have a function as $y={{\left( \sin x \right)}^{x}}$ then find the value of $\dfrac{dy}{dx}$

Answer 1

We solve this problem first by applying the logarithm function on both sides to remove the power. We use the theorem that is
$\log {{a}^{b}}=b\log a$
Then we differentiate with respect to $'x'$ on both sides to get the value of $\dfrac{dy}{dx}$
For finding the value of $\dfrac{dy}{dx}$ we use the product rule and chain rule.
The product rule states that if $u,v$ are some functions then
$\dfrac{d}{dx}\left( u\times v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$
The chain rule says that
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right).{g}'\left( x \right)$

Complete step by step solution:
We are given that the equation of $x,y$ as
$y={{\left( \sin x \right)}^{x}}$
Now, let us apply logarithm function on both sides then we get
$\Rightarrow \log y=\log {{\left( \sin x \right)}^{x}}$
We know that the theorem of logarithm that is
$\log {{a}^{b}}=b\log a$

By using this theorem in above equation we get
$\Rightarrow \log y=x\log \left( \sin x \right)$
Now, by differentiating with respect to $'x'$ on both sides we get
$\Rightarrow \dfrac{d}{dx}\left( \log y \right)=\dfrac{d}{dx}\left( x.\log \left( \sin x \right) \right)$
We know that the product rule of differentiation that if $u,v$ are some functions then
$\dfrac{d}{dx}\left( u\times v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$
By using the product rule in above equation we get
$\Rightarrow \dfrac{d}{dx}\left( \log y \right)=x.\dfrac{d}{dx}\left( \log \left( \sin x \right) \right)+\log \left( \sin x \right)\dfrac{dx}{dx}.......equation(i)$
We know that the standard formula of differentiation that is
$\Rightarrow \dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$
We also know that the chain rule of differentiation that is
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right).{g}'\left( x \right)$
By using the chain rule and the standard formula of differentiation in equation (i) we get
$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=x\left( \dfrac{1}{\sin x}\dfrac{d}{dx}\left( \sin x \right) \right)+\log \left( \sin x \right)$
We know that the standard formula of differentiation that is
$\Rightarrow \dfrac{d}{dx}\left( \sin x \right)=\cos x$
By using the this result above equation we get
$\Rightarrow \dfrac{dy}{dx}=y\left[ \dfrac{x}{\sin x}\left( \cos x \right)+\log \left( \sin x \right) \right]$
Now, by substituting $y={{\left( \sin x \right)}^{x}}$ in above equation we get
$\Rightarrow \dfrac{dy}{dx}={{\left( \sin x \right)}^{x}}\left[ x\cot x+\log \left( \sin x \right) \right]$
Therefore the value of $\dfrac{dy}{dx}$ is given as
$\therefore \dfrac{dy}{dx}={{\left( \sin x \right)}^{x}}\left[ x\cot x+\log \left( \sin x \right) \right]$

Note: We can solve this problem in another method.
We are given that the equation of $x,y$ as
$y={{\left( \sin x \right)}^{x}}$
Now, by differentiating with respect to $'x'$ on both sides we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left[ {{\left( \sin x \right)}^{x}} \right]$
If $u,v$ are the two functions then the general formula of differentiation is given as
$\dfrac{d}{dx}\left( {{u}^{v}} \right)=\dfrac{d}{dx}\left( {{u}^{v}}\text{ assuming }u\text{ as constant} \right)+\dfrac{d}{dx}\left( {{u}^{v}}\text{ assuming }v\text{ as constant} \right)$
By using the above formula we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{\left( \sin x \right)}^{x}}\text{assuming }\sin x\text{ as constant} \right)+\dfrac{d}{dx}\left( {{\left( \sin x \right)}^{x}}\text{assuming }x\text{ as constant} \right)$
Now we know that the general formula of differentiation that is
$\dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}\log a$ Where $'a'$ is constant.
By using this formula and chain rule to above equation we get

& \Rightarrow \dfrac{dy}{dx}={{\left( \sin x \right)}^{x}}\log \left( \sin x \right)+x.{{\left( \sin x \right)}^{x-1}}.\cos x \\\ & \Rightarrow \dfrac{dy}{dx}={{\left( \sin x \right)}^{x}}\log \left( \sin x \right)+x.{{\left( \sin x \right)}^{x}}.\cot x \\\ \end{aligned}$$ Therefore, by taking the common term out we get $$\therefore \dfrac{dy}{dx}={{\left( \sin x \right)}^{x}}\left[ x\cot x+\log \left( \sin x \right) \right]$$