How do you find the derivative of y=tan^2(x^4)?

just the chain rule

y = u^2 where u = tan(v) and v = x^4
y' = 2u u'
y' = 2u sec^2(v) v'
= 2u sec^2(v) (4x^3)
= 8x^3 tan(x^4) sec^2(x^4)

and, as usual, you can massage trig functions in various ways.

dy/dx = 2tan (x^4) * sec^2 (x^4) * 4x^3

To find the derivative of the function y = tan^2(x^4), we can use the chain rule. Let's go through the steps to find the derivative:

Step 1: Rewrite the function in a more convenient form.
y = tan^2(x^4) = (tan(x^4))^2

Step 2: Apply the chain rule.
The derivative of f(g(x))^n, where f(x) = x^2 and g(x) = tan(x), is given by:
f'(g(x)) * g'(x) * n * f(g(x))^(n-1)

In our case, n = 2, f(x) = x^2, and g(x) = tan(x^4).

Step 3: Find f'(g(x)) and g'(x).
f'(x) = 2x
g'(x) = sec^2(x^4) * (4x^3) [using the chain rule for g(x) = tan(x^4)]

Step 4: Substitute the derivatives into the chain rule formula.
y' = f'(g(x)) * g'(x) * n * f(g(x))^(n-1)
= 2(tan(x^4))^1 * sec^2(x^4) * (4x^3) * 2(tan(x^4))^0
= 4x^3 * sec^2(x^4)

Therefore, the derivative of y = tan^2(x^4) is y' = 4x^3 * sec^2(x^4).