A lot of time we have seen functions defined as $y=f(x)$. This clearly shows that $y$ is a function of the independent variable $x$. But often functions are defined implicitly. For instance, consider the equation $x^2+y^2=25$. Of course this is the equation of circle centered at the center $(0,0)$ with radius $5$. Also circles are not functions. But if we say $y\geq 0$, then the equation describes the upper half-circle which is a function defined by $y=\sqrt{25-x^2}$. Functions defined by equations like $x^2+y^2=25$ are called *implicit functions*. In some cases like $x^2+y^2=25$, we can easily write an implicit function explicitly as $y=f(x)$, but in many cases we cannot. For example, $x^3+y^3=6xy$. So, we need to devise a way to differentiate an implicit function without writing it as $y=f(x)$. This can indeed be done by the chain rule. You just assume that $y$ is a function of $x$ and use the chain rule. For example,

\begin{align*}

\frac{d}{dx}y^n&=(y^n)’\frac{dy}{dx}\ (y\ \mbox{is the innermost function})\\

&=ny^{n-1}\frac{dy}{dx}.

\end{align*}

Let us take a look at another example.

\begin{align*}

\frac{d}{dx}\cos y&=(\cos y)’\frac{dy}{dx}\ (y\ \mbox{is the innermost function})\\

&=-\sin y\frac{dy}{dx}.

\end{align*}

Here come more examples.

*Example*. If $x^2+y^2=25$, find $\frac{dy}{dx}$.

*Solution*. Differentiating the equation with respect to $x$, we obtain

$$2x+2y\frac{dy}{dx}=0.$$

Solving the resulting equation for $\frac{dy}{dx}$, we obtain

$$\frac{dy}{dx}=-\frac{x}{y}.$$

*Example*.

1. Find $y’$ if $x^3+y^3=6xy$.

*Solution*. Differentiate the equation with respect to $x$. Then we obtain

$$3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.$$

Solving the resulting equation for $\frac{dy}{dx}$, we obtain

$$\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}.$$

2. Find the tangent to $x^3+y^3=6xy$ at $(3,3)$.

*Solution*. The equation of tangent is

$$y-3=\left[\frac{dy}{dx}\right]_{(3,3)}(x-3).$$

$$\left[\frac{dy}{dx}\right]_{(3,3)}=\frac{2\cdot 3-(3)^2}{3^2-2\cdot 3}=-1.$$ Therefore, the tangent is given by $y=-x+6$.

robertahow do you go from

y−3=[dy/dx](3,3)(x−3).

to

[dy/dx](3,3)=2⋅3−(3)232−2⋅3=−1.

Sung LeePost authorIn the example part 1, we found $\frac{dy}{dx}$.

robertahow do you go from x2+y2=25 to 2x+2ydydx=0?

i feel there is a necessary step in there that is left out and make it harder for me to comprehend.

Sung LeePost authorI said “Differentiating the equation (meant $x^2+y^2=25$) with respect to $x$”. Maybe you are not understanding implicit differentiation. Recall the formula $\frac{d}{dx}y^n=ny^{n-1}\frac{dy}{dx}$ I discussed above.