Question

Find the derivative of the trigonometric function:
5 sec x + 4 cos x

Answer 1

Hint: First of all take y = 5 sec x + 4 cos x and assume 5 sec x = u and 4 cos x = v. Now, differentiate the equation and write $\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$. Now, use $\dfrac{d}{d\theta }\left( \sec \theta \right)=\sec \theta \tan \theta \text{ and }\dfrac{d}{d\theta }\left( \cos \theta \right)=-\sin \theta$ to finally get the required answers.

Complete step-by-step answer:
In this question, we have to find the derivative of 5 sec x + 4 cos x. Let us consider the expression given in the question.
y = 5 sec x + 4 cos x
Let us consider 5 sec x = u and 4 cos x = v. So, we get,
y = u + v
By differentiating both the sides of the above equation, we get,
$\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( 5\sec x \right)+\dfrac{dv}{dx}\left( 4\cos x \right)$
We know that $\dfrac{d}{d\theta }\left( \sec \theta \right)=\sec \theta \tan \theta$. By using this in the above equation, we get,
$\dfrac{dy}{dx}=5\sec x\tan x+\dfrac{d}{dx}\left( 4\cos x \right)$
We also know that $\dfrac{d}{d\theta }\left( \cos \theta \right)=-\sin \theta$. By using this in the above equation, we get,
$\dfrac{dy}{dx}=5\sec x\tan x+\left( -4\sin x \right)$
$\dfrac{dy}{dx}=5\sec x\tan x-4\sin x$
So, we get the derivative of 5 sec x + 4 cos x as 5 sec x tan x – 4 sin x.

Note: In these types of questions, where one function is a combination of various functions, it is advisable to differentiate each function separately. Also, students should memorize the differentiation of common functions like trigonometric functions, algebraic functions, etc. to easily solve the questions. Sometimes, students make this mistake of writing the differentiation of cos x as + sin x while actually, it is – sin x. So, this must be taken care of.