Differentiate:

A) y = tan^3 (5x-4)
B) y= 1/5 sin^5x -1/7sin^7 x
C) y = ln ((1+sin x)/(cos x))
D) y= cot x

When differentiating trigonometric functions it's less confusing to write them out fully.

e.g. rewrite $\tan^3 (5x-4)$ as $(\tan (5x-4))^3$.

It's easy to see that you can just differentiate this with the chain rule. There are three nested functions in that particular example so you will need to use the chain rule with three functions.

For D), you can either rewrite the function as $\frac{\cos x}{\sin x}$ or $(\tan x)^{-1}$ and proceed from there.

This site (https://www.derivative-calculator.net/) is very helpful for full solutions.

so basically think about it like this

sin^2(x) is same as (sin(x))^2

Use chain rule here for d/dx
d/dx(sin^2(x)) = 2(sinx)^1(cosx)
= 2sinxcosx

btw if possible can you bracket your answer better, ill solve them for you if you want further help,

You should use more brackets to make things clearer but for these questions remember how sec, cosec and cot are the reciprocals of sin, cos and tan respectively and you treat them as powers of -1. As for everything else remember your basic and trigonometric differentiation techniques. As for ln, you follow f'(x)/f(x) and simplify accordingly with your other identities such as tan=sin/cos and the reciprocals.

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•