Using Implicit Differentation to find dy/dx for a curve?

Question

I seem to be having trouble grasping the use of Implicit Differentiation. For example, one of the practice problems in my Calculus book states Use Implicit Differentiation to find $dy/dx$ for the curve $xy+sin y=x^2$.

Can someone walk me through step by step on how to properly do this? I'm completely lost.

Answer 1

You have to treat $y$ as a function of $x$. So, we go through the process term by term:

$(xy)'=\frac{d(x)}{dx}y+x\frac{dy}{dx}=y+x\frac{dy}{dx}$ from the product rule.

$(siny)'=cos(y)\frac{dy}{dx}$ from the chain rule.

$x^2=2x$, because we are simply taking the derivative of $x^2$. This gives us:

$$y+x\frac{dy}{dx}+cos(y)\frac{dy}{dx} =2x\implies \frac{dy}{dx}(x+cos(y))=2x-y\implies \frac{dy}{dx}=\frac{2x-y}{x+cos(y)}$$

The idea here is that some equations are defined so that $"y"$ is difficult to isolate. Implicit differentiation gives us an easier method to find the derivative of the function with respect to some variable $x$. (NOTE: As mentioned above: $(x+cos(y))$ must not be zero for this to hold.)

Answer 2

Both sides are functions of $x$, so you can differentiate them with respect to $x$. The derivative of the LHS is $y + xy' + y' \cos y$ (by the chain rule). The derivative of the RHS is $2x$. Putting this together we get

$$y + xy' + y' \cos y = 2x \implies y'(x + \cos y) = 2x - y \implies y' = \frac{2x - y}{x + \cos y}.$$

Answer 3

Explicit would mean solving for $y$ first, before differentiating. Implicit means we keep $y$ in its equation.

Just differentiating the equation regarding $x$ gives: $$ y + x y' + \cos(y) y' = y + (x + \cos(y)) y' = 2x $$ This is a new equation in $x$, $y$ and $y'$. Solving for $y'$ can be performed now, if wished for.

Answer 4

we assume that $xy(x)+\sin(y(x))=x^2$ then we get by differentiating with respect to $x$ $$y(x)+xy'(x)+\cos(y(x))y'(x)=2x$$