Slope of line tangent to function at point.

Question

I hope the picture's quality isn't too bad, but the question was slope of line tangent to $y^2 + (xy +1) ^3 = 0$ at point $(2,-1)$.

I tried two ways where one was trying to find the derivative right away- is this way wrong? especially the second part where I put $2(xy + 1) ^2$ ?

The second way I tried was actually multiplying it out because I was sort of hopeless. Bot hare wrong answers, and the slope was $\frac34$.

Please help me.

Answer 1

Function: $y^2 + (xy +1) ^3 = 0$

Differentiate implicitly: \begin{align*} y^2 + (xy +1) ^3 &= 0\\ \implies \frac{d}{dx}y^2 + (xy +1) ^3 &= 0\\ \implies 2y\frac{dy}{dx}+3(xy+1)^2\left(y+x\frac{dy}{dx}\right)&=0 \end{align*}

Now substitute $(x,y)$ and $\dfrac{dy}{dx}$ the subject.

Alternatively, you could make $y$ the subject in your original function.

Key Fact:

$$\frac{d}{dx}y^2=\frac{dy}{dx}\frac{d}{dy}y^2=\frac{dy}{dx}2y$$

Edit (Chain rule and implicit differentiation): \begin{align*} \frac{d}{dx} (xy +1) ^3 &= 0\\ 3(xy+1)^2\times\frac{d}{dx}(xy+1)&= 0 \end{align*} Note that for the chain rule, you just bring down the power from outside the bracket and reduce the power by one then perform the differentiation on the inside.