Implicit Differentiation

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,
\frac{d}{dx}y^n&=(y^n)’\frac{dy}{dx}\ (y\ \mbox{is the innermost function})\\
Let us take a look at another example.
\frac{d}{dx}\cos y&=(\cos y)’\frac{dy}{dx}\ (y\ \mbox{is the innermost function})\\
&=-\sin y\frac{dy}{dx}.
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
Solving the resulting equation for $\frac{dy}{dx}$, we obtain


1. Find $y’$ if $x^3+y^3=6xy$.

Solution. Differentiate the equation with respect to $x$. Then we obtain
Solving the resulting equation for $\frac{dy}{dx}$, we obtain

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

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

4 thoughts on “Implicit Differentiation

  1. roberta

    how 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.

    1. Sung Lee Post author

      I 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.


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