Question

Find gradient

xe^y + 4 ln y = x² at (1, 1)​

Answer 1

`xe^y+4\ln y=x^2`

Differentiate both sides with respect to x, assuming y = y(x).

`(\mathrm d(xe^y+4\ln y))/(\mathrm dx)=(\mathrm d(x^2))/(\mathrm dx)`

`(\mathrm d(xe^y))/(\mathrm dx)+(\mathrm d(4\ln y))/(\mathrm dx)=2x`

`(\mathrm d(x))/(\mathrm dx)e^y+x(\mathrm d(e^y))/(\mathrm dx)+\frac4y(\mathrm dy)/(\mathrm dx)=2x`

`e^y+xe^y(\mathrm dy)/(\mathrm dx)+\frac4y(\mathrm dy)/(\mathrm dx)=2x`

Solve for dy/dx :

`e^y+\left(xe^y+\frac4y\right)(\mathrm dy)/(\mathrm dx)=2x`

`\left(xe^y+\frac4y\right)(\mathrm dy)/(\mathrm dx)=2x-e^y`

`(\mathrm dy)/(\mathrm dx)=(2x-e^y)/(xe^y+\frac4y)`

If y ≠ 0, we can write

`(\mathrm dy)/(\mathrm dx)=(2xy-ye^y)/(xye^y+4)`

At the point (1, 1), the derivative is

`(\mathrm dy)/(\mathrm dx)\bigg|_(x=1,y=1)=\boxed{(2-e)/(e+4)}`