Question

Differentiate the given function \[{{\left( 5x \right)}^{3\cos 2x}}\].

Answer 1

Hint: In the above question of differentiation, first of all we will assume the given expression in a variable. Then we will take a logarithm function on both the sides of the equation. Now we will use the product rule as well as the chain rule of differentiation which is as follows:
Product rule: \[\dfrac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right]=f\left( x \right)g'\left( x \right)+g\left( x \right)f'\left( x \right)\]
Chain rule: \[\dfrac{d}{dx}\left[ f\left( g\left( h \right) \right) \right]=f'\left( g\left( h \right) \right)\times g'\left( h \right)\times h'\left( h \right)\times ...\]

Complete step-by-step solution -
We have been given the trigonometric function \[{{\left( 5x \right)}^{3\cos 2x}}\].
Let us assume the function to be y.
\[y={{\left( 5x \right)}^{3\cos 2x}}\]
Now taking logarithm function on both the sides of the equations we get as follows:
\[\log y=\log {{\left( 5x \right)}^{3\cos 2x}}\]
We already know that \[\log {{a}^{b}}=b\log a\].
So, by using this logarithmic formula in the above equation, we get as follows:
\[\log y=3\cos 2x.\log \left( 5x \right)\]
Now, differentiating both the sides of the equation with respect to x, we get as follows:
\[\dfrac{d}{dx}\left( \log y \right)=\dfrac{d}{dx}\left[ 3\cos 2x.\log \left( 5x \right) \right]\]
We already know that \[\dfrac{d}{dx}\log x=\dfrac{1}{x}\].
\[\dfrac{1}{y}.\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ 3\cos 2x.\log \left( 5x \right) \right]\]
Now using the product rule of differentiation, i.e. \[\dfrac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right]=f\left( x \right)g'\left( x \right)+g\left( x \right)f'\left( x \right)\] we get as follows:
\[\dfrac{1}{y}.\dfrac{dy}{dx}=3\left[ \cos 2x\dfrac{d}{dx}\left( \log 5x \right)+\left( \log 5x \right)\dfrac{d}{dx}\cos 2x \right]\]
We know the chain rule of differentiation, i.e. \[\dfrac{d}{dx}\left[ f\left( g\left( h \right) \right) \right]=f'\left( g\left( h \right) \right)\times g'\left( h \right)\times h'\left( h \right)\times ...\]
So by using the chain rule of differentiation, we get as follows:
\[\begin{align}
  & \dfrac{1}{y}.\dfrac{dy}{dx}=3\left[ \cos 2x\dfrac{1}{5x}\times \dfrac{d}{dx}\left( 5x \right)+\left( \log 5x \right)\left( -\sin 2x \right)\times \dfrac{d}{dx}2x \right] \\
 & \dfrac{1}{y}.\dfrac{dy}{dx}=3\left[ \dfrac{\cos 2x}{5x}\times 5-\sin 2x\times \log 5x\times 2 \right] \\
 & \dfrac{dy}{dx}=3y\left[ \dfrac{\cos 2x}{x}-2\sin 2x.\log 5x \right] \\
\end{align}\]
On substituting the value of ‘y’ in the above equation, we get as follows:
\[\dfrac{dy}{dx}=3\left( 5{{x}^{3\cos 2x}} \right)\left[ \dfrac{\cos 2x}{x}-2\sin 2x\left( \log 5x \right) \right]\]
Therefore, we get the differentiation of the given function to be equal to \[3\left( 5{{x}^{3\cos 2x}} \right)\left[ \dfrac{\cos 2x}{x}-2\sin 2x\left( \log 5x \right) \right]\].

Note: Be careful while taking logarithmic function to both the sides of the equation and using the logarithmic formula. Don’t miss that we have been given cos2x. Sometimes we put cosx instead of cos2x. Also, be careful while using the chain rule of differentiation as there is a chance of sign mistakes.