Question

What is the derivative of `y=e^(tan(x))`?

Answer 1

`e^tan(x)/cos^2(x)`

Explanation:

This is a composite function, i.e. a function composed by two functions `f(x)` and `g(x)`. The output of the inner function is used as input for the outer function, in the form `f(g(x))`.

In this case, the outer function is the exponential `e^x`, while the inner function is the tangen function `tan(x)`.

The differentiation of a composite function is governed by the chain rule:

`\frac{d}{dx} f(g(x)) = f'(g(x)) * g'(x)`

In other words, you derive the outer function, keeping the inner function as its input, and then multiply everything by the derivative of the inner function.

So, the derivative of the outer function, `e^x`, is still `e^x`, and we keep the inner function as input, so we have `e^tan(x)`.

Then, we multiply by the derivative of the inner function, and the derivative of `tan(x)` is `1/cos^2(x)`

So, we have

`\frac{d}{dx} e^tan(x) = e^tan(x)/cos^2(x)`

Answer 2

`(dy)/(dx)=e^(tanx)*(sec^2x)`

Explanation:

Use the general equation:

`d/(dx)(e^f(x))=f'(x)*e^f(x)`

In this case:
`e^f(x)=e^(tanx)`
and
`f'(x)=sec^2x`

As such:

`(dy)/(dx)=e^(tanx)*(sec^2x)`

For more information on how to differentiate trigonometric functions and `e`, you can refer to Khan Academy videos:

https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-advanced/ab-diff-exp/v/derivative-of-ex

https://www.khanacademy.org/math/ap-calculus-ab/ab-derivative-rules/ab-diff-trig-func/v/trig-functions-differentiation-sec

Good luck :)