Question

what would be the first step to get the derivative of y=1-cos2x+2cos^2x

Answer 1

Mmm so what are you stuck on?
The trig function got you confused?

Answer 2

trig functions*

Answer 3

We'll have to apply chain rule in each case.

Answer 4

so both are 2 different chain rules so then it would be 1-2cos2x but i dont know what it would be for 2cos^2x because of it's exponent on cos

Answer 5

Woops you applied the chain, the derivative of the inner function (2x).
But you never took the derivative of the outer function, cosine.

Answer 6

\[\Large\bf\sf \left(-\cos 2x\right)'\quad=\quad \sin2x\color{royalblue}{(2x)'}\quad=\quad 2 \sin 2x\]

Answer 7

The derivative of a `constant` is zero.
So the 1 differentiates to 0.

Answer 8

ok i see now, but the 2nd part still confuses me

Answer 9

It's confusing because of the power notation we use on trig functions.
It makes it difficult to see what the `outermost` function truly is.\[\Large\bf\sf 2\cos^2x\quad=\quad 2(\cos x)^2\]From this form we can see that we need to apply the power rule to the outer function first.

Answer 10

\[\Large\bf\sf \frac{d}{dx}2(\cos x)^2\quad=\quad 2\cdot 2(\cos x)^1\color{royalblue}{\frac{d}{dx}\cos x}\]

Answer 11

Understand that first step in differentiating it?
I applied the power rule to the 2 exponent.
The blue part is due to the chain rule, we need to find the derivative of that still.

Answer 12

Now I see, it would be best to just rewrite it like that, thanks so much!