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

Jun 7, 2018

$u ' = \cos \left({e}^{t}\right) {e}^{t}$

#### Explanation:

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

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:

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

In your case, $f \left(x\right) = \sin \left(x\right)$, and $g \left(x\right) = {e}^{x}$

So, since $f ' \left(x\right) = \cos \left(x\right)$, you have $f ' \left(g \left(x\right)\right) = \cos \left({e}^{t}\right)$

And since $g \left(x\right) = {e}^{x}$, you have $g ' \left(x\right) = g \left(x\right) = {e}^{x}$.

So, the expression becomes

$f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right) = \cos \left({e}^{x}\right) \cdot {e}^{x}$

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

$u ' = \cos \left({e}^{t}\right) {e}^{t}$