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(\mathrm dy)/(\mathrm dx)-\cos(y+4x)\left((\mathrm dy)/(\mathrm dx)+4\right)+\frac1{1+x}=0`

`2x\cos y-4\cos(y+4x)-\left(x^2\sin y+\cos(y+4x)\right)(\mathrm dy)/(\mathrm dx)+\frac1{1+x}=0`

`\left(x^2\sin y+\cos(y+4x)\right)(\mathrm dy)/(\mathrm dx)=2x\cos y-4\cos(y+4x)+\frac1{1+x}`

`(\mathrm dy)/(\mathrm dx)=\frac{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

`(\mathrm dy)/(\mathrm dx)(0,0)=(0-4\cos0+1)/(0+\cos0)=-3`