# Find the derivative of s = t^2 - cos t + 5e^t.

## Question:

Find the derivative of {eq}\; s = t^2 - \cos t + 5e^t {/eq}.

## Sum Rule for Derivatives:

Assume that the function {eq}s(t) {/eq} is equal to the sum of the functions {eq}f(t) {/eq}, {eq}g(t) {/eq} and {eq}h(t) {/eq}.

To find its derivative, we employ the sum rule for derivatives:

{eq}\displaystyle \frac{\mathrm{d}s}{\mathrm{d}t} = f'(t) + g'(t) + h'(t) {/eq}

This particular rule simply tells us that the derivative of a sum must be equivalent to the sum of the derivatives.

We employ the sum rule for derivatives to find the derivative of the given function, which is a sum of three functions:

{eq}\begin{align*} s & = t^2 - \cos t + 5e^t \\ \displaystyle \frac{\mathrm{d}s}{\mathrm{d}t} & =D_t( t^2 - \cos t + 5e^t )\\ \displaystyle \frac{\mathrm{d}s}{\mathrm{d}t} & =D_t( t^2) - D_t(\cos t )+D_t( 5e^t ) \;\;\; \left[ \mathrm{ Sum \ Rule \ for \ Derivatives }\right]\\ \displaystyle \frac{\mathrm{d}s}{\mathrm{d}t} & = 2t - (- \sin t) + 5e^t\\ \implies \displaystyle \frac{\mathrm{d}s}{\mathrm{d}t} & = 2t +\sin t + 5e^t\\ \end{align*} {/eq}