Question

How do you differentiate $ y={{\sec }^{2}}x+{{\tan }^{2}}x $ ? \[\]

Answer 1

Hint: We recall the sum rule of differentiation $ \dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right) $ .We recall the chain rule of differentiation $ \dfrac{dw}{dx}=\dfrac{dw}{du}\times \dfrac{du}{dx} $ where we first take $ w=gof={{\sec }^{2}}x $ and $ u=f\left( x \right)=\sec x $ . We first find $ u=f\left( x \right) $ as the function inside the bracket and $ w={{\tan }^{2}}x $ as the composite function and then differentiate using chain rule and the known differentiation $ \dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x $ . We similarly proceed for $ y=gof={{\tan }^{2}}x,v=\tan x $ using the known differentiation $ \dfrac{d}{dx}\tan x={{\sec }^{2}}x $ .\[\]

Complete step-by-step answer:
We know that If $ f\left( x \right),g\left( x \right) $ are two real valued functions the differentiation of their sum is given by the sum rule as
\[\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)\]

We know that if the functions $ f\left( x \right),g\left( x \right) $ are well defined real valued function 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 given the following function to differentiate
\[y={{\sec }^{2}}x+{{\tan }^{2}}x\]
We differentiate both sides of the above equation using sum rule to have;
 \[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{\sec }^{2}}x+{{\tan }^{2}}x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}{{\sec }^{2}}x+\dfrac{d}{dx}{{\tan }^{2}}x \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}{{\left( \sec x \right)}^{2}}+\dfrac{d}{dx}{{\left( \tan x \right)}^{2}}......\left( 1 \right) \\
\end{align}\]
We are asked to differentiate the function $ {{\sec }^{2}}x={{\left( \sec x \right)}^{2}} $ . We see that it is a composite function made by functions polynomial function that is $ {{x}^{2}} $ and trigonometric function that is $ \sec x $ . Let us assign the function within the bracket as $ f\left( x \right)=\sec x=u $ and $ g\left( x \right)={{x}^{2}} $ . So we have $ g\left( f\left( x \right) \right)=g\left( \sec x \right)={{\left( \sec x \right)}^{2}}=w $ . We differentiate using chain rule to have;
\[\begin{align}
  & \dfrac{dw}{dx}=\dfrac{dw}{du}\times \dfrac{du}{dx} \\
 & \Rightarrow \dfrac{d}{dx}w=\dfrac{d}{du}w\times \dfrac{d}{dx}u \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \sec x \right)}^{2}}=\dfrac{d}{d\left( \sec x \right)}{{\left( \sec x \right)}^{n}}\times \dfrac{d}{dx}\left( \sec x \right) \\
\end{align}\]
We know that from standard differentiation of polynomial function as $ \dfrac{d}{dt}{{t}^{n}}=n{{t}^{n-1}} $ for $ t=\sec x $ and standard differentiation of $ \sec x $ that is $ \dfrac{d}{dx}\sec x=\sec x\tan x $ to have;
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}{{\left( \sec x \right)}^{2}}={{\left( \sec x \right)}^{2-1}}\times \sec x\tan x \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \sec x \right)}^{2}}=2\sec x\times \sec x\tan x \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \sec x \right)}^{2}}=2{{\sec }^{2}}x\tan x.....\left( 2 \right) \\
\end{align}\]
We similarly differentiate $ {{\tan }^{2}}x $ by taking $ v=\tan x $ . We take $ f\left( x \right)=\tan x=v,g\left( x \right)={{x}^{2}} $ to have $ gof\left( x \right)={{\left( \tan x \right)}^{2}}=w $ . We differentiate using chain rule to have;
 \[\begin{align}
  & \dfrac{dw}{dx}=\dfrac{dw}{dv}\times \dfrac{dv}{dx} \\
 & \Rightarrow \dfrac{d}{dx}w=\dfrac{d}{dv}w\times \dfrac{d}{dx}v \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \tan x \right)}^{2}}=\dfrac{d}{d\left( \tan x \right)}{{\left( \tan x \right)}^{2}}\times \dfrac{d}{dx}\left( \tan x \right) \\
\end{align}\]
We know that from standard differentiation of polynomial function $ \dfrac{d}{dt}{{t}^{n}}=n{{t}^{n-1}} $ for $ t=\tan x $ and standard differentiation of $ \tan x $ that is $ \dfrac{d}{dx}\tan x={{\sec }^{2}}x $ to have;
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}{{\left( \tan x \right)}^{2}}={{\left( \tan x \right)}^{2-1}}\times {{\sec }^{2}}x \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \tan x \right)}^{2}}=2\tan x\times {{\sec }^{2}}x \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \tan x \right)}^{2}}=2{{\sec }^{2}}x\tan x.......\left( 3 \right) \\
\end{align}\]
We put (2) and (3) in (1) to have the result as
\[\begin{align}
  & \dfrac{dy}{dx}=2{{\sec }^{2}}x\tan x+2{{\sec }^{2}}x\tan x \\
 & \Rightarrow \dfrac{dy}{dx}=4{{\sec }^{2}}x\tan x \\
\end{align}\]

Note: We note that the given two trigonometric functions $ \sec x,\tan x $ are not defined for $ x=\left( 2n+1 \right)\dfrac{\pi }{2} $ . So we cannot get derivative of the given $ y={{\sec }^{2}}x+{{\tan }^{2}}x $ function at $ x=\left( 2n+1 \right)\dfrac{\pi }{2} $ . We should not try to convert $ {{\sec }^{2}}x,{{\tan }^{2}}x $ in to sine and cosine to differentiate because that will be long and difficult. If we are given $ y={{\sec }^{2}}x-{{\tan }^{2}}x $ it derivative will be zero since $ {{\sec }^{2}}x-{{\tan }^{2}}x=1 $