Question

How do you find the derivative of `y=tan^2(5x)` ?

Answer 1

The derivative of tan2(5x) is 10tan(5x)sec2(5x).

Explanation:

The derivative of `tan^2(5x)` is `10tan(5x)sec^2(5x)`.

Our function is the composite of three simpler functions:
Start with `x` and multipy it by `5`.
Then find the tangent of that.
Finally, find the square of that.

So to find the derivative we need the threefold chain rule:

If `k(x) = f(g(h(x)))`, then
`k'(x) = f'((g(h(x))*g'(h(x))*h'(x)`.

In our problem, `f(x) = x^2, g(x) = tan(x)`, and `h(x) = 5x`,
so `f'(x) = 2x, g'(x) = sec^2(x)` and `h'(x) = 5`

Thus `k'(x) = f'((g(h(x))*g'(h(x))*h'(x)`
`= f'(tan5x)*g'(5x)*h'(x)`
`=2tan(5x)*sec^2(5x)*5`
`=10tan(5x)sec^2(5x)`

Note: You might prefer doing the problem by letting
`u=5x, v=tan u, y = v^2`

Then
`dy/dx = ((dy)/(dv))((dv)/(du))((du)/(dx))`
`=2v*sec^2u*5`
`=10tan(5x)sec^2(5x)`

Many students prefer the second method, but occasionally you may encounter a problem for which the notation in the second method becomes ambiguous, so it is good to know both methods.