# Calculate the derivative of  y = (x^2+2)^2(x^4+4)^4  using logarithms?

Nov 22, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(4 x\right) \left(5 {x}^{4} + 8 {x}^{2} + 4\right) \left({x}^{2} + 2\right) {\left({x}^{4} + 4\right)}^{3}$

#### Explanation:

$y = {\left({x}^{2} + 2\right)}^{2} {\left({x}^{4} + 4\right)}^{4}$

Taking Natural Logarithms:

$\ln y = \ln \left\{{\left({x}^{2} + 2\right)}^{2} {\left({x}^{4} + 4\right)}^{4}\right\}$
$\therefore \ln y = \ln {\left({x}^{2} + 2\right)}^{2} + \ln {\left({x}^{4} + 4\right)}^{4}$
$\therefore \ln y = 2 \ln \left({x}^{2} + 2\right) + 4 \ln \left({x}^{4} + 4\right)$

Differentiating we get:
$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \frac{2 x}{{x}^{2} + 2} + 4 \frac{4 {x}^{3}}{{x}^{4} + 4}$

Simplifying;
$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(4 x\right) \left\{\frac{1}{{x}^{2} + 2} + \frac{4 {x}^{2}}{{x}^{4} + 4}\right\}$
$\therefore \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(4 x\right) \left\{\frac{\left({x}^{4} + 4\right) + 4 {x}^{2} \left({x}^{2} + 2\right)}{\left({x}^{2} + 2\right) \left({x}^{4} + 4\right)}\right\}$
$\therefore \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(4 x\right) \left\{\frac{{x}^{4} + 4 + 4 {x}^{4} + 8 {x}^{2}}{\left({x}^{2} + 2\right) \left({x}^{4} + 4\right)}\right\}$
$\therefore \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(4 x\right) \left\{\frac{5 {x}^{4} + 8 {x}^{2} + 4}{\left({x}^{2} + 2\right) \left({x}^{4} + 4\right)}\right\}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left(4 x\right) \left\{\frac{5 {x}^{4} + 8 {x}^{2} + 4}{\left({x}^{2} + 2\right) \left({x}^{4} + 4\right)}\right\} y$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left(4 x\right) \left\{\frac{5 {x}^{4} + 8 {x}^{2} + 4}{\left({x}^{2} + 2\right) \left({x}^{4} + 4\right)}\right\} {\left({x}^{2} + 2\right)}^{2} {\left({x}^{4} + 4\right)}^{4}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left(4 x\right) \left(5 {x}^{4} + 8 {x}^{2} + 4\right) \left({x}^{2} + 2\right) {\left({x}^{4} + 4\right)}^{3}$