Find y' if y=tan^3(x^2)?

Question

I tried to use the chain rule to do this problem, but I can't figure out how to differentiate the outside term tan(u)^3. It should be 3sec^2(u)tan^2(u) but I can't figure out how to get it. Thanks.

Answer 1

Think of tan^3 (x^2) as [tan(x^2)]^3

Using the chain rule:
y' = 3[tan(x^2)]^(3-1) * [tan(x^2)]'
Remember that the derivative of tangent is secant squared.
y' = 3[tan(x^2)]^2 * [sec(x^2)]^2 * (x^2)'
y' = 3[tan(x^2)]^2 * [sec(x^2)]^2 * (2x)
y' = 6x * [tan(x^2)]^2 * [sec(x^2)]^2

Answer 2

The part you asked about:

f(z) = z^3
g(z) = tan(z)
Let u=x^2:

y' = f'(g(u)) * g'(u)
y' = 3tan^2(u) * sec^2(u)

f', where f=z^3, is 3z^2.
g', where g=tan(z), is sec^2(z)

The full answer:

f(z) = z^3
g(z) = tan(z)
h(z) = z^2

y = f(g(h(x)))
y' = f'(g(h(x))) * g'
y' = f'(g(h(x))) * g'(h(x)) * h'(x)
y' = 3tan^2(x^2) * sec^2(x^2) * 2x
y' = 6x * tan^2(x^2) * sec^2(x^2)

Answer 3

y=tan^3(x^2)
Put x^2 = u, du = 2x dx.
y = tan^3(u) =[ tan(u) ]^3
Use the chain rule again.
Put v = tan(u), dv = sec^2(u)du.
y = v^3
dy/dx = (dy / dv)(dv /du)(du / dx)
= (3v^2) (sec^2(u)) (2x)
= 3tan^2(u)sec^2(u)(2x)
= 3tan^2(x^2)sec^2(x^2) (2x)
= 6x tan^2(x^2) sec^2(x^2).