Question

What are the derivatives of $\sec 2x$ and $\tan 2x$?

Answer 1

Hint: First consider the function $\sec 2x$. Assume ‘h’ as the small change in x and hence find the small change in the function f (x) given as f(x + h). Now, use the limit definition of derivative and apply the formula $f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$, substitute the value of the given functions. Use the conversion $\sec x=\dfrac{1}{\cos x}$ and apply the formula $\left( \cos a-\cos b \right)=-2\sin \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right)$ to simplify. Use the basic limit formula $\displaystyle \lim_{h \to 0}\dfrac{\sin h}{h}=1$ to get the answer. Now, consider the function $\tan x$ and apply the same approach of limit definition of the derivative. Use the conversion $\tan x=\dfrac{\sin x}{\cos x}$ then use the trigonometric identity $\left( \sin a\cos b-\cos a\sin b \right)=\sin \left( a-b \right)$. Again, use the formula $\displaystyle \lim_{h \to 0}\dfrac{\sin h}{h}=1$ to get the answer.

Complete step by step answer:
Here we have been provided with the function $\sec 2x$ and $\tan 2x$. We are asked to find their derivative. Let us use the limit definition of derivative also known as the first principle to get the answer.
We know that derivative of a function is defined as the rate of change of function. In other words we can say that it is the measure of change in the value of the function with respect to the change in the value of the variable on which it depends. This change is infinitesimally small, that means tending to zero. Mathematically we have,
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$
In the above formula we have ‘h’ as the infinitesimally small change in the variable x.
(1) Let us consider the function $\sec 2x$ first. Assuming the function $f\left( x \right)=\sec 2x$, so substituting $\left( x+h \right)$ in place of x in the function, we get,
$\Rightarrow f\left( x+h \right)=\sec 2\left( x+h \right)$
Now, substituting the value of the functions f (x) and f (x + h) in the limit formula we get,
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{\sec 2\left( x+h \right)-\sec 2x}{h} \right)$
Using the conversion $\sec x=\dfrac{1}{\cos x}$ we get,
$\begin{align}
  & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{\dfrac{1}{\cos 2\left( x+h \right)}-\dfrac{1}{\cos 2x}}{h} \right) \\
 & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{\cos 2x-\cos 2\left( x+h \right)}{h\cos 2\left( x+h \right)\cos 2x} \right) \\
\end{align}$
Using the formula $\left( \cos a-\cos b \right)=-2\sin \left( \dfrac{a+b}{2} \right)\sin \left( \dfrac{a-b}{2} \right)$ we get,
$\begin{align}
  & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{-2\sin \left( \dfrac{2x+2h+2x}{2} \right)\sin \left( \dfrac{2x-2x-2h}{2} \right)}{h\cos 2\left( x+h \right)\cos 2x} \right) \\
 & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{-2\sin \left( 2x+h \right)\sin \left( -h \right)}{h\cos 2\left( x+h \right)\cos 2x} \right) \\
 & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{2\sin \left( 2x+h \right)}{\cos 2\left( x+h \right)\cos 2x} \right)\times \displaystyle \lim_{h \to 0}\dfrac{\sin h}{h} \\
\end{align}$
Using the basic formula of limit $\displaystyle \lim_{h \to 0}\dfrac{\sin h}{h}=1$ and substituting h = 0 we get,
\[\begin{align}
  & \Rightarrow f'\left( x \right)=\left( \dfrac{2\sin \left( 2x \right)}{\cos 2\left( x \right)\cos 2x} \right)\times 1 \\
 & \Rightarrow f'\left( x \right)=\left( \dfrac{2\sin \left( 2x \right)}{\cos 2\left( x \right)\cos 2x} \right) \\
 & \therefore f'\left( x \right)=2\tan 2x\sec 2x \\
\end{align}\]
Hence, the derivative of $\sec 2x$ is $\left( 2\tan 2x\sec 2x \right)$.
(2) Let us consider the function $\tan 2x$ first. Assuming the function $f\left( x \right)=\tan 2x$, so substituting $\left( x+h \right)$ in place of x in the function we get,
$\Rightarrow f\left( x+h \right)=\tan 2\left( x+h \right)$
Now, substituting the value of the functions f (x) and f (x + h) in the limit formula, we get,
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{\tan 2\left( x+h \right)-\tan 2x}{h} \right)$
Using the conversion $\tan x=\dfrac{\sin x}{\cos x}$ we get,
$\begin{align}
  & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{\dfrac{\sin 2\left( x+h \right)}{\cos 2\left( x+h \right)}-\dfrac{\sin 2x}{\cos 2x}}{h} \right) \\
 & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{\sin 2\left( x+h \right)\cos 2x-\cos 2\left( x+h \right)\sin 2x}{h\cos 2\left( x+h \right)\cos 2x} \right) \\
\end{align}$
Using the formula $\left( \sin a\cos b-\cos a\sin b \right)=\sin \left( a-b \right)$ we get,
$\begin{align}
  & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{sin\left( 2x+2h-2x \right)}{h\cos 2\left( x+h \right)\cos 2x} \right) \\
 & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{\sin \left( 2h \right)}{h\cos 2\left( x+h \right)\cos 2x} \right) \\
 & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{1}{\cos 2\left( x+h \right)\cos 2x} \right)\times \displaystyle \lim_{h \to 0}\dfrac{\sin 2h}{h} \\
\end{align}$
Using the identity $\sin 2a=2\sin a\cos a$ we get,
$\begin{align}
  & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{1}{\cos 2\left( x+h \right)\cos 2x} \right)\times \displaystyle \lim_{h \to 0}\dfrac{2\sin h\cos h}{h} \\
\end{align}$
Again, using the basic formula of limit $\displaystyle \lim_{h \to 0}\dfrac{\sin h}{h}=1$ and substituting h = 0 we get,
\[\begin{align}
  & \Rightarrow f'\left( x \right)=\left( \dfrac{1}{\cos 2\left( x \right)\cos 2x} \right)\times 2 \\
 & \Rightarrow f'\left( x \right)=\left( \dfrac{2}{{{\cos }^{2}}2x} \right) \\
 & \therefore f'\left( x \right)=2{{\sec }^{2}}2x \\
\end{align}\]

Hence, the derivative of $\tan 2x$ is $\left( 2{{\sec }^{2}}2x \right)$.

Note: Note that all the formulas of derivatives of different functions are derived from the first principle. You have to remember the formulas of all the trigonometric functions as they will not be derived everywhere using the first principle. Here, if you know the derivatives of $\tan x$ and $\sec x$ then you can easily apply the chain rule of derivatives to get the answer. According to this rule first you have to consider the derivative of the given functions with respect to their argument (2x) and then the derivative of 2x with respect to x. Finally both the derivatives will be multiplied to get the answer.