Question

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

Answer 1

Hint: 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.