How can I prove d/dx (sin 5x) = 5cos 5x knowing that d/5dx (sin 5x) = cos 5x?

Answer 1

substitute u = 5x, we obtain y = sin u

remember the chain rule for sin u!

d(sin u) / dx = cos u * du/dx

since u = 5x, du/dx = 5

now substitute du/dx in the chain rule and you have your answer...

d(sin u) / dx = cos u * 5 = 5 * cos (5x)...

Answer 2

you have to know the chain rule:

f'(g(x))(g'(x):

fprime of function g times gprime

(this is one of the established laws of derivatives
but if you require a proof it is probably somewhere on the internet)

f(x)=sinx g(x)=5x

f'=cosx g'=5

therefore

therefore fprime(5x) times gprime

=cos(5x)*5

=5cos(5x)

Answer 3

y = sin u

dy/ dx = d (sin u) /dx

d (sin u) /dx = (cos u)*(du/dx).

If u = 5x,

du/dx = 5* dx/dx = 5.

Therefore, dy/ dx = (cos u)*5

= 5 cos 5x.

Answer 4

first of all answer sin 5x.
Solve sin5x = cos5x
then differentiate 5x=5
therfore it is equal to 5cos5x.

Answer 5

We can prove it by following way;
d/dx(sin5x)=d/5dx(sin5x)x5
=cos5x

Answer 6

wat in the world is d/5dx (sin 5x)?

anyway, just mutiply noth side by 5 and its proven :)