Question

What is the derivative of `y=sec^2(x) + tan^2(x)`?

Answer 1

The derivative of `y=sec^2x + tan^2x` is:

`4sec^2xtanx`

Process:

Since the derivative of a sum is equal to the sum of the derivatives, we can just derive `sec^2x` and `tan^2x` separately and add them together.

For the derivative of `sec^2x`, we must apply the Chain Rule:

`F(x) = f(g(x))`
`F'(x) = f'(g(x))g'(x)`,

with the outer function being `x^2`, and the inner function being `secx`. Now we find the derivative of the outer function while keeping the inner function the same, then multiply it by the derivative of the inner function. This gives us:

`f(x) = x^2`
`f'(x) = 2x`

`g(x) = secx`
`g'(x) = secxtanx`

Plugging these into our Chain Rule formula, we have:

`F'(x) = f'(g(x))g'(x)`,
`F'(x) = 2(secx)secxtanx = 2sec^2xtanx`

Now we follow the same process for the `tan^2x` term, replacing `secx` with `tanx`, ending up with:

`f(x) = x^2`
`f'(x) = 2x`

`g(x) = tanx`
`g'(x) = sec^2x`

`F'(x) = f'(g(x))g'(x)`,
`F'(x) = 2(tanx)sec^2x = 2sec^2xtanx`

Adding these terms together, we have our final answer:

`2sec^2xtanx + 2sec^2xtanx`

= `4sec^2xtanx`