Question

If the given expression is $y={{x}^{x}}+{{\left( \sin x \right)}^{\cot x}}$. Find $\dfrac{dy}{dx}$.

Answer 1

Here first of all take ${{x}^{x}}=u$ and ${{\left( \sin x \right)}^{\cot x}}=v$ to get $\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$. Now to get $\dfrac{dy}{dx}$, separately differentiate u and v and substitute it in the expression of $\dfrac{dy}{dx}$. Also, for $u\left( x \right)={{\left[ f\left( x \right) \right]}^{g\left( x \right)}}$, always first take log on both sides and then differentiate.

Complete step-by-step answer :
We are given that $y={{x}^{x}}+{{\left( \sin x \right)}^{\cot x}}$. We have to find the value of $\dfrac{dy}{dx}$.
Let us first consider the function given in the question.
$y={{x}^{x}}+{{\left( \sin x \right)}^{\cot x}}$
Here, let us assume ${{x}^{x}}=u$ and ${{\left( \sin x \right)}^{\cot x}}=v$
So, we get, y = u + v
By differentiating both sides of the above equation with respect to x, we get,
$\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}....\left( i \right)$
Let us consider the function ${{x}^{x}}$, that is
$u={{x}^{x}}$
By taking log on both sides, we get,
$\log u=\log {{x}^{x}}$
We know that $\log {{m}^{n}}=n\log m$. By applying this in the RHS of the above equation, we get,
$\log u=x\log x$
Now, by differentiating both the sides with respect to x, we get,
$\dfrac{d}{dx}\left( \log u \right)=\dfrac{d}{dx}\left( x\log x \right)$
We know that, by chain rule if y = f (u), then $\dfrac{dy}{dx}={{f}^{'}}\left( u \right)\dfrac{du}{dx}$
Also, we know that $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$. By applying these in the LHS of the above equation, we get,
$\dfrac{1}{u}\dfrac{du}{dx}=\dfrac{d}{dx}\left( x\log x \right)$
We know that by product rule $\dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)=f.\dfrac{dg}{dx}+g.\dfrac{df}{dx}$
By applying this in the RHS of the above equation, we get,
$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=x\dfrac{d}{dx}\left( \log x \right)+\log x\dfrac{d}{dx}\left( x \right)$
$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=x.\left( \dfrac{1}{x} \right)+\log x.\left( 1 \right)$
By multiplying ‘u’ on both sides, we get,
$\dfrac{du}{dx}=u\left( 1+\log x \right)$
By substituting the value of $u={{x}^{x}}$. We get,
$\dfrac{du}{dx}={{x}^{x}}\left( 1+\log x \right)....\left( ii \right)$
Now, let us consider the function ${{\left( \sin x \right)}^{\cot x}}$, that is
$v={{\left( \sin x \right)}^{\cot x}}$
By taking log on both sides, we get,
$\log v=\log {{\left( \sin x \right)}^{\cot x}}$
We know that $\log {{m}^{n}}=n\log m$. By applying this in the RHS of the above equation, we get,
$\log v=\cot x.\log \left( \sin x \right)$
Now, by differentiating both sides with respect to x, we get,
$\dfrac{d}{dx}\left( \log v \right)=\dfrac{d}{dx}\left[ \cot x.\log \left( \sin x \right) \right]$
Now, by applying chain rule in LHS of the above equation, we get,
$\dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{d}{dx}\left[ \cot x.\log \left( \sin x \right) \right]$
Now, by applying product rule in RHS of the above equation, we get,
$\dfrac{1}{v}\dfrac{dv}{dx}=\cot x.\dfrac{d}{dx}\left[ \log \left( \sin x \right) \right]+\left[ \log \left( \sin x \right) \right].\dfrac{d}{dx}\left( \cot x \right)$
By applying chain rule in the RHS of the above equation, we get,
$\dfrac{1}{v}\dfrac{dv}{dx}=\left( \cot x \right).\left( \dfrac{1}{\sin x} \right)\dfrac{d}{dx}\left( \sin x \right)+\left[ \log \left( \sin x \right) \right].\dfrac{d}{dx}\left( \cot x \right)$
We know that $\dfrac{d}{dx}\left( \sin x \right)=\cos x$ and $\dfrac{d}{dx}\left( \cot x \right)=\left( -{{\operatorname{cosec}}^{2}}x \right)$.
By applying these in the above equation, we get,
$\dfrac{1}{v}.\dfrac{dv}{dx}=\dfrac{\left( \cot x \right)}{\sin x}.\left( \cos x \right)+\left[ \log \left( \sin x \right) \right]\left( -{{\operatorname{cosec}}^{2}}x \right)$
Now, by multiplying ‘v’ on both sides, we get,
$\dfrac{dv}{dx}=v\left( \dfrac{\cot x}{\sin x}.\cos x-\left( {{\operatorname{cosec}}^{2}}x \right)\log \left( \sin x \right) \right)$
By substituting $v=\log {{\left( \sin x \right)}^{\cot x}}$ and $\dfrac{\cos x}{\sin x}=\cot x$, we get,
$\dfrac{dv}{dx}=\log {{\left( \sin x \right)}^{\cot x}}\left[ {{\left( \cot x \right)}^{2}}-\left( {{\operatorname{cosec}}^{2}}x \right).\log \left( \sin x \right) \right]$
Or, $\dfrac{dv}{dx}=\left( \cot x \right)\log \left( \sin x \right)\left[ {{\left( \cot x \right)}^{2}}-\left( {{\operatorname{cosec}}^{2}}x \right)\log \left( \sin x \right) \right]$
By taking $\left( \cot x \right)\log \left( \sin x \right)$ inside the bracket, we get,
$\dfrac{dv}{dx}={{\left( \cot x \right)}^{3}}\log \left( \sin x \right)-\left( {{\operatorname{cosec}}^{2}}x \right)\left( \cot x \right){{\left[ \log \left( \sin x \right) \right]}^{2}}....\left( iii \right)$
Now by substituting values of $\dfrac{du}{dx}$ and $\dfrac{dv}{dx}$ from equation (ii) and (iii) in equation (i), we get,
$\dfrac{dy}{dx}={{x}^{x}}\left( 1+\log x \right)+\left[ {{\left( \cot x \right)}^{3}}\log \left( \sin x \right)-\left( {{\operatorname{cosec}}^{2}}x \right)\left( \cot x \right){{\left( \log \left( \sin x \right) \right)}^{2}} \right]$
Therefore, we have got the value of $\dfrac{dy}{dx}$.

Note :Students are advised to always take the functions in parts and then differentiate it to avoid confusion in case of large functions. Also, many students make this mistake of calculating $\dfrac{d}{dx}\left( \log v \right)=\dfrac{1}{v}$ and $\dfrac{d}{dx}\left( \log u \right)=\dfrac{1}{u}$ but this is wrong because according to the chain rule, $\dfrac{d}{dx}\left( \log v \right)=\dfrac{1}{v}.\dfrac{dv}{dx}$ and $\dfrac{d}{dx}\left( \log u \right)=\dfrac{1}{u}.\dfrac{du}{dx}$. So this mistake must be avoided.