Question

Find the derivative, do not simplify;

y = CubeRoot(cos(2x))

I convert the problem to;

y=cos(2x)^(1/3) and I know that the derivative of cos(x) = -sin(x), but I am unsure if anything special needs to be done with the 2 and the 1/3 exponent. Thank you!

Answer 1

you need to use the chain rule

taking the 1/3 in front getting and subtracting 1
(1/3cos(2x)^-2/3)
then do the cos inner function
-sin(2x)
finally the inner inner function 2x
2
getting the final answer of
(1/3cos(2x)^-2/3)(-sin(2x))(2)

Answer 2

Chain rule:\[\frac{d}{dx} f(g(x)) = f'(g(x))g'(x).\]Let\[f(x) = x^{1/3}\]\[g(x) = \cos{x}\]\[h(x) = 2x,\]then\[y = f(g(h(x)))\]and using the chain rule\[y' = f'(g(h(x)))(g(h(x)))'\]using the chain rule again:\[y' = f'(g(h(x)))g'(h(x))h'(x).\]Also,\[f'(x) = \frac{1}{3}x^{-2/3}\]\[g'(x) = -\sin{x}\]\[h'(x) = 2\]so\[y' = \frac{1}{3}(g(h(x)))^{-2/3}\cdot (-\sin(h(x)))\cdot 2 = -\frac{2}{3}(\cos(2x))^{-2/3}\sin(2x) = -\frac{2}{3}\frac{\sin(2x)}{\cos^{2/3}(2x)}.\]

Answer 3

lol well he simplified his so i would go with his :) plus it looks better

Answer 4

Thanks a lot, guys!