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Question: What is the derivative of \(y={{\sec }^{2}}x?\)...

What is the derivative of y=sec⁡2x?y={{\sec }^{2}}x?

Explanation

Solution

If we need to find the derivative of the function fn(x),{{f}^{n}}\left( x \right), we will first differentiate this function for the exponent and then we will differentiate the function. Mathematically, we write ddxfn(x)=nfn−1(x)f′(x).\dfrac{d}{dx}{{f}^{n}}\left( x \right)=n{{f}^{n-1}}\left( x \right){f}'\left( x \right). The derivative of sec⁡x\sec x is sec⁡xtan⁡x.\sec x\tan x.

Complete step by step solution:
Let us consider the given problem.
We are asked to find the derivative of the given function.
The given function we need to differentiate is y=sec⁡2x.y={{\sec }^{2}}x.
Before start differentiating the given function, we should know the rule of differentiation given by ddxfn(x)=nfn−1(x)f′(x).\dfrac{d}{dx}{{f}^{n}}\left( x \right)=n{{f}^{n-1}}\left( x \right){f}'\left( x \right).
So, we will differentiate the given function considering that the given function is of the form xn.{{x}^{n}}. Then, we will differentiate the function regardless of the exponent.
So, we can write dydx=ddxsec⁡2x.\dfrac{dy}{dx}=\dfrac{d}{dx}{{\sec }^{2}}x.
Also, we should know that the derivative of sec⁡x\sec x is sec⁡xtan⁡x.\sec x\tan x.
Let us suppose that f(x)=sec⁡x.f\left( x \right)=\sec x. Then, we will get f2(x)=sec⁡2x.{{f}^{2}}\left( x \right)={{\sec }^{2}}x.
We will get, 2f2−1(x)=2f(x)=2sec⁡x.2{{f}^{2-1}}\left( x \right)=2f\left( x \right)=2\sec x.
Similarly, we will get f′(x)=sec⁡xtan⁡x.{f}'\left( x \right)=\sec x\tan x.
Now the first part of the derivative will be 2sec⁡x.2\sec x.
The second part of the derivative will be sec⁡xtan⁡x.\sec x\tan x.
That is, we will get dydx=ddxsec⁡2x=2sec⁡xsec⁡xtan⁡x.\dfrac{dy}{dx}=\dfrac{d}{dx}{{\sec }^{2}}x=2\sec x\sec x\tan x.
That is, dydx=ddxsec⁡2x=2sec⁡xsec⁡xtan⁡x=2sec⁡2xtan⁡x.\dfrac{dy}{dx}=\dfrac{d}{dx}{{\sec }^{2}}x=2\sec x\sec x\tan x=2{{\sec }^{2}}x\tan x.
Since y=sec⁡2x,y={{\sec }^{2}}x, we will substitute this in the above equation.
So, we will get the derivative of the given function as dydx=2sec⁡2xtan⁡x=2ytan⁡x.\dfrac{dy}{dx}=2{{\sec }^{2}}x\tan x=2y\tan x.
Hence the derivative of the given function is obtained as dydx=2ytan⁡x.\dfrac{dy}{dx}=2y\tan x.

Note: Let us recall the derivatives of the basic trigonometric functions. The derivative of sin⁡x\sin x is cos⁡x.\cos x. The derivative of cos⁡x\cos x is −sin⁡x.-\sin x. The derivative of tan⁡x\tan x is sec⁡2x.{{\sec }^{2}}x. The derivative of sec⁡x\sec x is sec⁡xtan⁡x.\sec x\tan x. The derivative of cot⁡x\cot x is −cos⁡ec2x.-\cos e{{c}^{2}}x. The derivative of cos⁡ecx\cos ecx is −cos⁡ecxcot⁡x.-\cos ecx\cot x. When we differentiate a function, we should first consider the exponent, if any.