Question

Help Please! Attachment to come.

Answer 1

Do you want to know how to differentiate these with the chain rule? or just want the answers?

Answer 2

i would like to know how to differentiate them with the chain rule please!

Answer 3

These two should be memorised. \[\frac{ d(uv) }{ dx } = u \frac{ dv }{ dx } + v \frac{ du }{ dx }\] \[\frac{ dy }{dx } = \frac{ dy }{ dt } \frac{ dt }{ dx }\]
For the first question,using the chain rule we let y=cos(x)^3 , cos(x)=u
first, differentiate y with respect to u.
\[y=\cos^3(x) = u^3\]
\[\frac{ dy }{ du } = 3u^2\]
then we differentiate u with respect to x:
\[u=\cos(x)\]
\[\frac{ du }{ dx } = -\sin x\]
by chain rule,
\[\frac{ dy }{dx } = \frac{ dy }{ du } \frac{ du }{ dx }\]
\[\frac{ dy }{dx } = (3 u^2 )(-\sin x)\]
substituting cos x for u,
\[\frac{ dy }{ dx } =- 3 \cos^2 x \sin x\]
to get the answer C, we:
\[- 3 \cos^2 x \frac{ \sin x \cos x }{ \cos x } = -3 \cos^3 x \frac{ \sin x }{\cos x } = -3 \cos^3 x \tan x\]

Answer 4

@Shadowys and @Zekarias thanks so much for putting so much detail into your solution! can you show me how to do the other too??

Answer 5

lol I'll show you how to do the second one. The others are simply rinse and repeat.
let y= sin x tan x
\[\frac{ dy }{ dx } = \sin x \left [ \frac{ d(\tan x) }{dx } \right ] + \tan x \left [ \frac{ d(\sin x) }{ dx} \right]\]
= \[\sin x \sec^2 x + \tan x \cos x\]
simplifying,
= \[\tan x \sec x + \sin x \], which is D

Answer 6

Alright! thank you very much! you were very helpful! :D

Answer 7

You're welcome~ :)

Answer 8

I hope you will finish the last one.