what did i do wrong

[ocean prairie]

im guessing i just plugged into the multiplication rule wrong or i didnt do my derivatives of each term being multiplied right

[ocean prairie]

i did f'=ncos^(n-1) (x)*-sin^n (x) and g'=-sin(nx) times (n) (i cant do an asterisk again cuz it makes it a different format) and then fg'+gf'

sry if thats hard to read idk how to do the texit thing or whatever its called

[dusk bane]

Chain rule!

@ocean prairie

[halcyon pumice]

Both your f' and g' have chain rule slightly wrong

[cyan plume]

It is a product of two terms. It has been a while for me but a product uses the product rule where you add, not subtract, the derivative of term 2 and the product of term 1 with the product of term 2 with the derivative of term 1. It looks like you subtracted those two products.

[dusk bane]

derivative of cos(x) is -sin(x) 👍

I think that’s where the negative is coming from

[cyan plume]

That explains the negative yes. But how do I explain the subtraction of the next term? Should it not be adding it?

[dusk bane]

Distributing the - sign?

Throughout the second term

It makes it subtraction

Adding a negative is same thing as subtracting a positive

[cyan plume]

If it is f’(x)g(x) + g’(x)f(x), we do not distribute. The negative lives in the f’(x).

[dusk bane]

The original function h(x) contains a product of cosines

…

Both f’(x) and g’(x) will have a negative term

[cyan plume]

Okay that would be right if we put in brackets around the whole thing and placed the - in front of the first bracket.

[dusk bane]

I don’t think you understand what I’m saying

[cyan plume]

Sorry I guess I do not.

I worked it out by hand and I see both terms need a negative. Maybe that is what you meant.

What my answer has is an “n” in the first term. This is because the derivative of cos(nx) is -sin(nx)n.

[dusk bane]

Yea that’s what I mean, both terms contain a negative so when you add them, it’s a negative + (negative) which is the same as negative - (positive)

[dapper marsh]

Ima write this down

Well Javier already did it so maybe not

[dapper marsh]

This is right I’m pretty sure

[cyan plume]

I think the original post had a math calculator like Mathematica that said his answer (minus the “n” in the first term) was incorrect and if he put the suggested answer in there it would reject or accept it right away.

[ocean prairie]

oh shoot i forgot to put the n at the front of my equation i just saw that is that all i missed

cuz i had the n at the front when i wrote it down and also when i did g' in my first message (i said times n cuz i couldnt do another astersik and i didnt want to just put parenthesis even though that would work)

[dapper marsh]

The fact that’s all you got wrong is what sucks I feel ye I’m taking calc rn and I sometimes forget little things like that

But everything else was great

That shows you know what your doing

[ocean prairie]

Yeah I feel like taking it in school next year is going to be way easier now that I’m doing most of calc 1 out of school and learning more about integrals and calc 2 outside of that class as well (it’s a math class that has consistently been around 3 years ahead of school but only last year and this year slowed down to 1 year before)

[jade scarab]

your first term should be $-nsin(nx)cos^n (x)$

[raven isle]

OK32

[jade scarab]

when you do $\frac{d}{dx}(cos(nx))$

[raven isle]

OK32

[jade scarab]

you were right to differentiate the cos to a negative sign, but now you have to also differentiate the inside of the function and multiply that

chain rule

[jade scarab]

oh oops you got that, mb

[carmine jungle]

there is another mistake the derivative of cos^nx is not -n.cos^(n-1)(x).sin^n(x)

you need to apply chain rule correctly

h(x) =f(g(x))
h'(x) = f'(g(x)).g'(x)

after you write derivative of the outer power function you multiply by derivate of cosx , (-sinx) you do not raise it to n

$\frac{d}{dx}(\cos^n(x)) = n\cos^{n-1}x(-\sin x)$