just think of it like this:
d/dx of sin(u) = u' cos(u)
d/dx sin(x) = 1(cos(x))
d/dx sin(4x) = 4(cos(4x))
d/dx sin(x/6) [or, sin((1/6)x)] = (1/6)(cos(x/6))
if F(x) = sin^5 (x/6) + cos^4 (x/3)
find F'(x)
I'm doing the wrong derivative, i thought that the derivative of
sin(x/6) = cos (x/6)
but that doesn't seem to be correct.
Let's say we're looking at
(sin(x/6))^5 ( i took the 5 out for chain rule purposes)
then
5(sin(x/6))^4 is the derivative of the outside applied to the inside,
and the derivative of the inside(sin(x/6)) is x/6 * (cos(x/6))??
i thought that the derivative would be
5(sin(x/6))^4 * (cos(x/6))
is there a rule for the x/6 to come out of there?
I can solve the problem, i just need to figure out if i'm missing something specifically with the derivative.
i can't exactly keep going if the derivative is wrong
thanks a bunch
And, just in case a picture helps...
Don't integrate - balloontegrate!
Balloon Calculus: worked examples from past papers