find f ' x for sin^3 x?

Question

i've tried it and i came up with 3 cos x , but i have no idea if this is correct , tell me if u come up with the same

Answer 1

It's 3*sin^2 (x) * cos(x)

To do it step by step:
Let u = sin(x)
du/dx = cos(x)
You have that part

f = u^3
df/du = 3*u^2

df/dx = df/fu * du/dx
= 3*u^2 * cos(x)
= 3*sin^2 (x) * cos(x)

You're forgetting to find df/du

Answer 2

f(x) = (sin x)^3
Use the chain rule.
f'(x) = 3(sin x)^2 * (cos x)
Basically you just do your normal power rule, bring the power down and lower the power, then multiply by the derivative of sin x, which is cos x.

A good way to remember this is the derivative of the outside evaluated at the inside times the derivative of the inside.

Answer 3

see simple dx^n/dx = n*x^(n-1)
if f(x) = x^4 then f'(x) = 4x^3

chain rule d [ ]^n / dx = n*[ ]^(n-1) *d [ ] /dx
example
if f(x) = (x^3)^4 ;; [ ] = x^3
then f'(x) = 4 [ x^3 ] ^3 * (3x^2)
= 12 x^11 that like f(x) = x^12 ,f'(x) = 12x^11

for f(x)=sin^3 x = [ ] ^3
f'(x) = 3 [ sinx ] ^2 *cosx
= 3(cosx)(sin^2x) ^-^

Answer 4

Let u = sinx, then du/dx = cos x
dy/dx = 3u^2du/dx
= 3sin^2xcosx

Answer 5

f(x)=sin^3x = sin x* sin^2x

d/dx(sinx)=cosx

d/dx sin^2= d/dx(sinx*sinx)

d/dx sin^x= cosxsinx+cosxsinx= 2cosxsinx

d/dx f(x) = cosx(sin^2x)+(2cosxsinx)sinx

f'(x)= 3cos x sin^x

Not garanteed to be correct.