Implicit Differentiation — Question 8

Question 8 - Solution

At $(1,1)$ both sides of the corrected equation equal $\ln2+1$.

(a) Implicit derivative

Differentiating (with $x+y>0$) gives

$$\frac{1+y'}{x+y}+2xy+x^2y'=2yy'+3.$$

Therefore

$$\boxed{y'=\frac{3-2xy-1/(x+y)}{1/(x+y)+x^2-2y}}.$$

(b) Tangent slope

At $(1,1)$ the numerator is $1/2$ and the denominator is $-1/2$, hence the slope is $\boxed{-1}$.