# How do I find f'(x) for f(x)=x^2*10^(2x) ?

Sep 20, 2014

The derivative of $f \left(x\right) = {x}^{2} \cdot {10}^{2 x}$ is

$f ' \left(x\right) = 2 x \left(1 + x \ln 10\right) {10}^{2 x}$

Let us look at some details.

We need the following tools in your toolbox.

1. Power Rule: $\left({x}^{n}\right) ' = n {x}^{n - 1}$

2. Exponential Rule: $\left({b}^{x}\right) ' = \left(\ln b\right) {b}^{x}$

3. Product Rule: $\left[f \left(x\right) \cdot g \left(x\right)\right] ' = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

4. Chain Rule: $\left[f \left(g \left(x\right)\right)\right] ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Let us find $\left({10}^{2 x}\right) '$ first.

By Chain Rule and Exponential Rule,

$\left({10}^{2 x}\right) ' = \left(\ln 10\right) {10}^{2 x} \cdot \left(2 x\right) ' = 2 \left(\ln 10\right) {10}^{2 x}$

Now, we can find $f ' \left(x\right)$.

By Product Rule,

$f ' \left(x\right) = \left({x}^{2}\right) ' \cdot {10}^{2 x} + {x}^{2} \cdot \left({10}^{2 x}\right) '$

by Power Rule and the derivative we found above,

$= 2 x \cdot {10}^{2 x} + {x}^{2} \cdot 2 \left(\ln 10\right) {10}^{2 x}$

by factoring out $2 x {10}^{2 x}$,

$= 2 x \left(1 + x \ln 10\right) {10}^{2 x}$