Question

How do you find the derivative of `u=sin(e^t)`?

Answer 1

`u' = cos(e^t)e^t`

Explanation:

This is an example of a composed function, i.e. take two function `f(x)` and `g(x)`, and use the output of one as the input for the other, like `f(g(x))` or `g(f(x))`.

There is a rule to differentiate such functions: first of all, derive the outer function, keeping the inner function as argument. Then, multiply by the derivative of the inner function. In formulas:

`d/(dx) f(g(x)) = f'(g(x)) * g'(x)`

In your case, `f(x) = sin(x)`, and `g(x) = e^x`

So, since `f'(x) = cos(x)`, you have `f'(g(x)) = cos(e^t)`

And since `g(x)=e^x`, you have `g'(x)=g(x)=e^x`.

So, the expression becomes

`f'(g(x)) * g'(x) = cos(e^x)*e^x`

Now, instead of the usual notation `y=f(x)` you were using `u=f(t)`, but this is totally irrelevant, since the name of the variables plays no role, so if you prefer your derivative is

`u' = cos(e^t)e^t`