Question

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

Answer 1

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