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\frac{d}{dx}\left(y x^2+\ln\left(2x+y\right)=0\right)

Find the derivative of yx^2+ln(2x+y)=0

Answer

$\frac{2}{y+2x}+2y\cdot x=0$

Step-by-step explanation

Problem

$\frac{d}{dx}\left(y x^2+\ln\left(2x+y\right)=0\right)$
1

Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable

$\frac{d}{dx}\left(\ln\left(y+2x\right)+x^2y\right)=\frac{d}{dx}\left(0\right)$

Unlock this step-by-step solution!

Answer

$\frac{2}{y+2x}+2y\cdot x=0$
$\frac{d}{dx}\left(y x^2+\ln\left(2x+y\right)=0\right)$

Main topic:

Differential calculus

Used formulas:

5. See formulas

Time to solve it:

0.25 seconds