Question

If we have a trigonometric function given by \[y=\tan x+\sec x\], then prove that:
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\operatorname{cosx}}{{{\left( 1-\operatorname{sinx} \right)}^{2}}}\]

Answer 1

Hint: First of all we will have to know about the \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]. Actually it means the double differentiation of the function ‘y’ with respect to ‘x’. So we will differentiate the function ‘y’ twice and arrange them to proceed the given expression using trigonometric identity. We can use the fact that the derivative of tanx is \[{{\sec }^{2}}x\] and of secx is secxtanx. Then we can proceed to find the second derivative.

Complete step-by-step answer:
We have been given \[y=\tan x+\sec x\] and asked to prove \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\operatorname{cosx}}{{{\left( 1-\operatorname{sinx} \right)}^{2}}}\].
We know that \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] is nothing but double differentiation of the function with respect to ‘x’, so we will differentiate ‘y’ twice.
\[\begin{align}
  & \Rightarrow y=\tan x+\sec x \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \tan x+\sec x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \tan x \right)}{dx}+\dfrac{d\left( \sec x \right)}{dx} \\
\end{align}\]
Since we know that \[\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x\] and \[\dfrac{d\left( \sec x \right)}{dx}=\sec x\tan x\]
\[\Rightarrow \dfrac{dy}{dx}={{\sec }^{2}}x+\sec x\tan x\]
Again, differentiating both sides of the equation, we get as follows:
\[\begin{align}
  & \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( {{\sec }^{2}}x+\sec x\tan x \right) \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( {{\sec }^{2}}x \right)+\dfrac{d}{dx}\left( \operatorname{secxtanx} \right) \\
\end{align}\]
Now for \[\dfrac{d}{dx}\left( {{\sec }^{2}}x \right)\] we will use the chain rule as follows:
\[\begin{align}
  & \dfrac{d}{dx}\left[ F\left( g\left( h\left( x \right) \right) \right) \right]=F'\left( g\left( h\left( x \right) \right) \right)\times g'\left( h\left( x \right) \right)\times h'\left( x \right)\times 1 \\
 & \Rightarrow \dfrac{d}{dx}\left( {{\sec }^{2}}x \right)=2\sec x\dfrac{d}{dx}\left( \sec x \right) \\
 & =2\sec x\times \sec x\tan x \\
 & =2{{\sec }^{2}}x\tan x \\
\end{align}\]
Now for \[\dfrac{d}{dx}\left( \operatorname{secxtanx} \right)\] we will use the product rule of differentiation as follows:
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( F\left( x \right).g\left( x \right) \right)=F\left( x \right)g'\left( x \right)+g\left( x \right)F'\left( x \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( \operatorname{secxtanx} \right)=\sec x\dfrac{d}{dx}\left( \tan x \right)+\tan x\dfrac{d}{dx}\left( \sec x \right) \\
 & =\sec x\times {{\sec }^{2}}x+\tan x\times \sec x\times \tan x \\
 & ={{\sec }^{3}}x+{{\tan }^{2}}x\sec x \\
\end{align}\]
Substituting the values of differentiation, we get as follows:
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2{{\sec }^{2}}\tan x+{{\sec }^{3}}x+{{\tan }^{2}}x\sec x \\
 & =2{{\sec }^{2}}x\tan x+\sec x\left( {{\sec }^{2}}x+{{\tan }^{2}}x \right) \\
\end{align}\]
Since we know that \[1+{{\tan }^{2}}x={{\sec }^{2}}x\]
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2{{\sec }^{2}}x\tan x+\sec x\left( 1+{{\tan }^{2}}x+{{\tan }^{2}}x \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2{{\sec }^{2}}x\tan x+\sec x\left( 1+2{{\tan }^{2}}x \right) \\
 & =2{{\sec }^{2}}x\tan x+\sec x+2{{\tan }^{2}}x\sec x \\
\end{align}\]
We know that \[\tan x=\dfrac{\sin x}{\cos x}\] and \[\sec x=\dfrac{1}{\cos x}\].
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\times \dfrac{1}{{{\cos }^{2}}x}\times \dfrac{\sin x}{\cos x}+\dfrac{1}{\cos x}+\dfrac{2{{\sin }^{2}}x}{{{\cos }^{2}}x}\times \dfrac{1}{\cos x} \\
 & =\dfrac{2\sin x}{{{\cos }^{3}}x}+\dfrac{1}{\cos x}+\dfrac{2{{\sin }^{2}}x}{{{\cos }^{3}}x} \\
\end{align}\]
Taking LCM of the terms we get as follows:
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2\sin x+{{\cos }^{2}}x+2{{\sin }^{2}}x}{{{\cos }^{3}}x}=\dfrac{2\sin x+2{{\sin }^{2}}x+1-{{\sin }^{2}}x}{\cos x.{{\cos }^{2}}x}\]
Since we know that \[{{\cos }^{2}}x=1-{{\sin }^{2}}x\]
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2\sin x\left( 1+\sin x \right)+\left( 1+\sin x \right)\left( 1-\sin x \right)}{\cos x\left( 1-{{\sin }^{2}}x \right)}=\dfrac{\left( 2\sin x+1-\sin x \right)\left( 1+\sin x \right)}{\cos x\left( 1+\sin x \right)\left( 1-\sin x \right)} \\
 & =\dfrac{\left( 1+\sin x \right)\left( 1+\sin x \right)}{\cos x\left( 1+\sin x \right)\left( 1-\sin x \right)}=\dfrac{\left( 1+\sin x \right)\left( 1-\sin x \right)}{\cos x\left( 1-\sin x \right)\left( 1-\sin x \right)} \\
\end{align}\]
Since we know that \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1-{{\sin }^{2}}x}{\cos x{{\left( 1-\sin x \right)}^{2}}} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{\cos }^{2}}x}{\cos x{{\left( 1-\sin x \right)}^{2}}} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\cos x}{{{\left( 1-\sin x \right)}^{2}}} \\
\end{align}\]
Hence it is proved.

Note: Be careful while doing differentiation and take care of the sign at each step of the calculation. We can also solve this by converting secx and tanx into sinx and cosx terms in the beginning and then finding its derivative. Also, remember that the first differentiation of ‘y’ with respect to x is denoted by \[\dfrac{dy}{dx}\] or y’ and second differentiation is denoted by \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] or y’’.