# How do I find f'(x) for f(x)=5^x ?

Apr 4, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = {5}^{x} \log 5$

#### Explanation:

$y = {5}^{x}$

$\log y = x \log 5$

$\frac{\mathrm{dy}}{\mathrm{dx}} . \frac{1}{y} = x . \left(0\right) + \log 5. \left(1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} . \frac{1}{y} = \log 5$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \log 5 . y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \log 5 . {5}^{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {5}^{x} \log 5$

Apr 4, 2018

$f \left(x\right) = y = {5}^{x}$

Taking log on both the sides,

$\log y = \log {5}^{x}$

$\log y = x \log 5$ color(white)(wwggggggggggggw $\left[\text{as } \textcolor{red}{\log {a}^{b} = b \log a}\right]$

Now, applying implicit differentiation,
$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = x \frac{d \left(\log 5\right)}{\mathrm{dx}} + \log 5$ color(white)(o $\left[\text{product rule is applied on } x \log 5\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left[x + \log 5\right]$ color(white)(gggggggw $\left[\text{as derivative of a constant is 1}\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {5}^{x} \left[x + \log 5\right]$

Apr 4, 2018

$f ' \left(x\right) = \ln 5 \cdot {5}^{x}$

#### Explanation:

Let $y = f \left(x\right) = {5}^{x}$

then $\ln y = x \ln 5$ and differentiating we get

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \ln 5$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \ln 5 \cdot y = \ln 5 \cdot {5}^{x}$

Apr 4, 2018

${5}^{x} \cdot \ln 5$

#### Explanation:

Given: $f \left(x\right) = {5}^{x}$

Using the common integral that $\left({a}^{x}\right) ' = {a}^{x} \ln a$, we get:

$f ' \left(x\right) = {5}^{x} \cdot \ln 5$