Question

How do you find the derivative of $\sec x\tan x$ ?

Answer 1

Hint: Here they have given a function to find the derivative of the same. So we can write the given function as $y = \sec x\tan x$ and now we need to find the derivative which is $\dfrac{{dy}}{{dx}}$ . But here we have two functions which are in the form of $u \times v$ form. Hence we can apply product rule that is $u \times v = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$ .

Complete step by step answer:
Here in this question, we have the function $\sec x\tan x$ for which we need to find the derivative that is $\dfrac{{dy}}{{dx}}$ . So we can write the function as $y = \sec x\tan x$ .
By looking at the given function we can notice that we have two functions that are $\sec x\tan x$ which represents like $u \times v$. Hence we can make use of the product rule of differentiation to solve the given problem. The product rule is given by: $u \times v = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$ . Here $u = \sec x$ and $v = \tan x$ .
The given function is, $y = \sec x\tan x$
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(\sec x\tan x)\]
Now, apply product rule to the above function, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \sec x \times \dfrac{d}{{dx}}(\tan x) + \tan x \times \dfrac{d}{{dx}}(\sec x)\]
We know that $\dfrac{d}{{dx}}(\tan x) = {\sec ^2}x$ and $\dfrac{d}{{dx}}(\sec x) = \sec x\tan x$ . now substitute these in the above expression, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \sec x \times {\sec ^2}x + \tan x \times \sec x\tan x\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \sec x \times {\sec ^2}x + \sec x{\tan ^2}x\]
We have $\sec x$ which is common in both the terms of the above expression, so take the common term outside for simplification purposes. Therefore, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \sec x\left( {{{\sec }^2}x + {{\tan }^2}x} \right)\]

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

Note: Here in this question, we have $\sec x\tan x$ which is a product of two functions, so we need to apply product rule as we did in the above procedure. While applying the product rule you should be careful otherwise you may get the wrong answer. Whenever we have trigonometric functions to find the derivative, we need to remember the formulas or the derivative of the basic (Basic functions are nothing but the sine, cosine, tangent, cosecant, secant, and the cotangent function) or standard functions to make the simplification easier.