Question

Find dy/dx at (0,0) if x^2cosy-sin(y+4x)+ln(1+x)=0

Answer 1

$x^2\cos y-\sin(y+4x)+\ln(1+x)=0$

Differentiate both sides with respect to $x$, treating $y$ as a function of $x$:

$2x\cos y-x^2\sin y\dfrac{\mathrm dy}{\mathrm dx}-\cos(y+4x)\left(\dfrac{\mathrm dy}{\mathrm dx}+4\right)+\dfrac1{1+x}=0$

$2x\cos y-4\cos(y+4x)-\left(x^2\sin y+\cos(y+4x)\right)\dfrac{\mathrm dy}{\mathrm dx}+\dfrac1{1+x}=0$

$\left(x^2\sin y+\cos(y+4x)\right)\dfrac{\mathrm dy}{\mathrm dx}=2x\cos y-4\cos(y+4x)+\dfrac1{1+x}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x\cos y-4\cos(y+4x)+\frac1{1+x}}{x^2\sin y+\cos(y+4x)}$

At the point (0, 0), the derivative is

$\dfrac{\mathrm dy}{\mathrm dx}(0,0)=\dfrac{0-4\cos0+1}{0+\cos0}=-3$