Implicit differentiation

Question

I have looked at several examples and I still fail at applying it to the question I have at hand. Youtube videos on the subject and even my math professor dont help.

My professor can't explain anything he says something ambiguous and follows it up with "it's easy".

I have 3 questions

use implicit differentiation to find $\frac{dy}{dx}$ in terms of $x$ and $y$

1.) $x^3-xy+y^2=4$

2.) $y=\sin(xy)$

find $\frac{d^2 y}{dx^2}$

1.) $x^2y^2 - 2x = 3$

Answer 1

I'll show you how to do 1), and you can try the others yourself. Differentiating implicitly, we have

$$\frac{d}{dx}(x^3 - xy + y^2) = \frac{d}{dx} 4$$ or $$3x^2 -y -xy' + 2yy' = 0$$ so that $$3x^2 - y = y'(x - 2y)$$ and hence $$y' = \frac{3x^2 - y}{x - 2y}.$$

Answer 2

And I'll show you 2, $$ dy=\cos(xy)*(ydx+xdy)=y\cos(xy)dx+x\cos(xy)dy $$ using chain and product rules. Then solve for $dy$, and then finally $\frac{dy}{dx}$, $$ dy-x\cos(xy)dy=\cos(xy)*(ydx+xdy)=y\cos(xy)dx\Rightarrow dy(1-x\cos(xy))=y\cos(xy)dx\Rightarrow \frac{dy}{dx}=\frac{y\cos(xy)}{1-x\cos(xy)} $$

Answer 3

Here is a method you may not have tried. See if it helps.

Take the derivative of both sides of the equation using "prime" notation making no assumption about which variable the derivative is being taken with respect to. Apply all derivative rules such as chain rule, product rule, etc.

\begin{eqnarray} \left(x^3-xy+y^2\right)^\prime&=&4^\prime\\ 3x^2x^\prime-(x^\prime y+xy^\prime)+2yy^\prime&=&0\\ 3x^2x^\prime- yx^\prime-xy^\prime+2yy^\prime&=&0 \end{eqnarray}

Next, replace every $x^\prime$ with $dx$ and every $y^\prime$ with $dy$ to obtain and factor those expressions out of the terms containing them.

\begin{eqnarray} 3x^2dx-y\,dx-x\,dy+2y\,dy&=&0\\ \left(3x^2-y\right)\,dx+(2y-x)\,dy&=&0 \end{eqnarray}

Now divide both sides by $dx$ and solve for $\frac{dy}{dx}$.

\begin{eqnarray} \left(3x^2-y\right)+(2y-x)\,\frac{dy}{dx}&=&0\\ \frac{dy}{dx}&=&\dfrac{y-3x^2}{2y-x} \end{eqnarray}