How do you differentiate 5^(2x^2)?

Oct 30, 2017

$\frac{d}{\mathrm{dx}} \left({5}^{2 {x}^{2}}\right) = 4 \ln 5 x {5}^{2 {x}^{2}}$

Explanation:

Consider that:

${5}^{2 {x}^{2}} = {\left({e}^{\ln} 5\right)}^{2 {x}^{2}} = {e}^{2 \ln 5 {x}^{2}}$

So, using the chain rule:

$\frac{d}{\mathrm{dx}} \left({5}^{2 {x}^{2}}\right) = \frac{d}{\mathrm{dx}} \left({e}^{2 \ln 5 {x}^{2}}\right) = {e}^{2 \ln 5 {x}^{2}} \frac{d}{\mathrm{dx}} \left(2 \ln 5 {x}^{2}\right)$

$\frac{d}{\mathrm{dx}} \left({5}^{2 {x}^{2}}\right) = \frac{d}{\mathrm{dx}} \left({e}^{2 \ln 5 {x}^{2}}\right) = 4 \ln 5 x {e}^{2 \ln 5 {x}^{2}} = 4 \ln 5 x {5}^{2 {x}^{2}}$

Oct 31, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = {5}^{2 {x}^{2}} 4 x \ln 5$

Explanation:

$y = {5}^{2 {x}^{2}}$

take natural logs of both sides

$\ln y = \ln {5}^{2 {x}^{2}}$

using laws of logs

$\ln y = 2 {x}^{2} \ln 5$

differentiate implicitly

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 4 x \ln 5$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y 4 x \ln 5$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {5}^{2 {x}^{2}} 4 x \ln 5$