Trigometric function of a function question.

Hi Guys,

I am having problems with the following question..

Evaluate dy(over)dx for y=7sin^6x

Here is where i am up to..

Let u = sinx > du(over)dx = cosx

y = 7u^6 > dy(over)du = 42u5

dy(over)dx = dy(over)du x du(over)dx = 42u^5x (cosx)

I now have to substitute U back?

Any help appreciated, thanks. :)

Re: Trigometric function of a function question.

Quote:

Originally Posted by

**imonlyhuman** Let u = sinx > du(over)dx = cosx

y = 7u^6 > dy(over)du = 42u5

dy(over)dx = dy(over)du x du(over)dx = 42u^5x (cosx)

I now have to substitute U back?

It's a little confusing when you use "x" as a multiplication sign and as a variable. But if I understand what you are saying, your work should be correct so far. Now you substitute,

Re: Trigometric function of a function question.

Thanks for your reply Reckoner,

Is this the last step?

Apologies, trigonometric substitution confuses me!

Re: Trigometric function of a function question.