How do you use implicit differentiation to find (dy)/(dx) given 3x^2+3=ln5xy^2?

Aug 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(3 x - \frac{1}{2 x}\right)$

Explanation:

Implicit differentiation is no different to normal, you just have to make sure to apply the chain rule when differentiating the $y$ terms, ie $y$ will differentiate to $\frac{\mathrm{dy}}{\mathrm{dx}}$.

$\frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 3 = \ln \left(5 x {y}^{2}\right)\right)$

When we differentiate the log I used a combination of chain and product rules, but you could also rewrite it as :

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

using rules of logs and then just use chain rule.

$6 x = \frac{1}{5 {y}^{2} x} \cdot \left(5 {y}^{2} + 10 x y \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$30 {x}^{2} {y}^{2} = 5 {y}^{2} + 10 x y \frac{\mathrm{dy}}{\mathrm{dx}}$

$5 {y}^{2} \left(6 {x}^{2} - 1\right) = 10 x y \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{y}{2} \left(6 x - \frac{1}{x}\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = y \left(3 x - \frac{1}{2 x}\right)$