Question

How do you find the derivative of $ y={{\tan }^{4}}\left( x \right) $ ?

Answer 1

Hint: We recall the definition of composite function $ gof\left( x \right)=g\left( f\left( x \right) \right) $ . We recall the chain rule of differentiation $ \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx} $ where $ y=gof={{\tan }^{4}}\left( x \right) $ and $ u=f\left( x \right)=\tan x $ . We first take $ u=f\left( x \right) $ as the function inside the bracket and $ y $ as the given function. We then differentiate using chain rule and the standard differentiation of $ \tan x $ that is $ \dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x $ .\[\]

Complete step-by-step answer:
We know from calculus that the derivative of a function of a real variable measures the rate of change of the functional value with respect to argument or input value. The process of finding derivative is called differentiation. If $ f\left( x \right) $ is real valued function then we use the differential operator $ \dfrac{d}{dx} $ and find the derivative as
\[\dfrac{d}{dx}f\left( x \right)={{f}^{'}}\left( x \right)\]
If the functions $ f\left( x \right),g\left( x \right) $ are defined within sets $ f:A\to B $ and $ g:B\to C $ then the composite function from A to C is defend as $ g\left( f\left( x \right) \right) $ within sets $ gof:A\to C $ . If we denote $ g\left( f\left( x \right) \right)=y $ and $ f\left( x \right)=u $ then we can differentiate the composite function using chain rule as
\[\dfrac{d}{dx}g\left( f\left( x \right) \right)=\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}\]
We are asked to differentiate the function $ {{\tan }^{4}}\left( x \right)={{\left( \tan x \right)}^{4}} $ . We see that it is a composite function made by functions polynomial function that is $ {{x}^{4}} $ and trigonometric function that is $ \tan x $ . Let us assign the function within the bracket as $ f\left( x \right)=\tan x=u $ and $ g\left( x \right)={{x}^{4}} $ . So we have $ g\left( f\left( x \right) \right)=g\left( \tan x \right)={{\left( \tan x \right)}^{4}}=y $ . We differentiate using chain rule to have;
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx} \\
 & \Rightarrow \dfrac{d}{dx}y=\dfrac{d}{du}y\times \dfrac{d}{dx}u \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \tan x \right)}^{4}}=\dfrac{d}{d\left( \tan x \right)}{{\left( \tan x \right)}^{4}}\times \dfrac{d}{dx}\left( \tan x \right) \\
\end{align}\]
We know that from standard differentiation of polynomial function as $ \dfrac{d}{dt}{{t}^{n}}=n{{t}^{n-1}} $ where $ n $ is any real number. We use it for $ t=\tan x,n=4 $ in the above step to have
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}{{\left( \tan x \right)}^{n}}=4{{\left( \tan x \right)}^{4-1}}\times \dfrac{d}{dx}\left( \tan x \right) \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \tan x \right)}^{n}}=4{{\left( \tan x \right)}^{3}}\times \dfrac{d}{dx}\left( \tan x \right) \\
\end{align}\]
We know that from standard differentiation of tangent function $ \dfrac{d}{dt}\tan t={{\sec }^{2}}t $ . We use it for $ t=x $ in the above step to have the derivative as
\[\Rightarrow \dfrac{d}{dx}{{\left( \tan x \right)}^{4}}=4{{\tan }^{3}} x {{\sec }^{2}}x\]

Note: We note that if one of the functions $ f\left( x \right),g\left( x \right) $ is not differentiable at some values of $ x $ then $ gof $ is also not differentiable at those values. Here $ \tan x $ does not exist for the values $ x=\left( 2n+1 \right)\dfrac{\pi }{2},n\in \mathsf{\mathbb{Z}} $ . Hence we cannot derivative of the given function $ {{\tan }^{4}}\left( x \right) $ at those values. We should also remember sum rule $ {{\left( f+g \right)}^{'}}={{f}^{'}}+{{g}^{'}} $ , the product rule $ {{\left( fg \right)}^{'}}=g{{f}^{'}}+f{{g}^{'}} $ and the quotient rule $ {{\left( \dfrac{f}{g} \right)}^{'}}=\dfrac{g{{f}^{'}}-f{{g}^{'}}}{{{g}^{2}}} $ of differentiation.