Question

What if the derivative of $y = {\sec ^2}(x)?$

Answer 1

Hint: First, the six basic trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent and all the fundamental trigonometric identities are derived from the six trigonometric ratios.
We need to analyze the given information so that we are able to solve the problem.
If we need to find the derivation of the given function ${f^n}(x)$, we will first differentiate this function for the exponent and then we will derive the function.
Formula used: The derivation of the function ${f^n}(x)$ is $\dfrac{d}{{dx}}{f^n}(x) = n{f^{n - 1}}(x){f^1}(x)$

Complete step by step answer:
From the given that we have$y = {\sec ^2}(x)$, we ask to find the derivation of the given function.
As we mention the derivation formula is ${f^n}(x)$is $\dfrac{d}{{dx}}{f^n}(x) = n{f^{n - 1}}(x){f^1}(x)$
So, we will differentiate the given function considering the given function as of the form \[{x^n}\]
Then, we will differentiate the function of the exponent, as $\dfrac{{dy}}{{dx}} = \dfrac{{d({{\sec }^2}(x))}}{{dx}}$
The derivative of the function $\sec x$is $\sec x\tan x$(as apply the derivative value)
Let us suppose $f(x) = \sec x$then we will get the square term as ${f^2}(x) = {\sec ^2}x$
Now, we will find the derivation as $2f(x) = 2\sec x$
Similarly, we can now proceed ${f^1}(x) = \sec x\tan x$
The derivation of the first part of the given function is $2\sec x$and the second derivation part is $secx\tan x$
Hence, we get, $\dfrac{{dy}}{{dx}} = \dfrac{{d({{\sec }^2}(x))}}{{dx}} = 2\sec x \times \sec x \times \tan x$
Further solving, we get $\dfrac{{dy}}{{dx}} = \dfrac{{d({{\sec }^2}(x))}}{{dx}} = 2{\sec ^2}x \times \tan x$
But since $y = {\sec ^2}(x)$applying this in above we get, $\dfrac{{dy}}{{dx}} = 2{\sec ^2}x \times \tan x \Rightarrow 2y\tan x$
Hence the derivation of the $y = {\sec ^2}(x)$is $2y\tan x$ or with converting that $y = {\sec ^2}(x)$, we can also say the derivation of the $y = {\sec ^2}(x)$is $2{\sec ^2}x \times \tan x$. The answer for the given question is both, before substituting and after substituting is the difference.

Note: The basic trigonometry derivatives are
$\cos x = - \sin x$
$\tan x = {\sec ^2}x$
$\sec x = \sec x\tan x$ (Which is used in the given problem)
${\cot ^2}x = - \cos e{c^2}x$
In the differentiation, we mostly use the chain rule for two or more two values that occur in the given composition.
The inverse process of the differentiation is integration like if the derivative of $\dfrac{{d({x^2})}}{{dx}} = 2x$ then the integration process of the same value is $\int {2x} dx = \dfrac{{2{x^2}}}{2} \Rightarrow {x^2}$
Hence both processes are inverse to each other, most likely the inverse operations multiplication and division.