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# Guide on Implicit Differentiation

schedule Aug 12, 2023
Last updated
local_offer
Calculus
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# Motivating example

In Calculus, we are often given a function $f(x)$ and asked to take the derivative with respect to $x$. For instance, given $f(x)=x^2$, we can easily obtain the derivative $f'(x)=2x$. However, what if the given equation is not a function of such forms? For instance, consider the equation of the circle:

$$\begin{equation}\label{eq:Fslf5tE3Tagbblgeb9s} x^2+y^2=3^2 \end{equation}$$

This is clearly not of form $f(x)$, but we can rearrange the equation to make $y$ the subject:

$$\begin{equation}\label{eq:kRqwKaeXzuUpPstSdgE} y=f(x)=\pm\sqrt{3^2-x^2} \end{equation}$$

Even though \eqref{eq:kRqwKaeXzuUpPstSdgE} is of form $f(x)$, the problem here is that we have a $\pm$, which means we actually have two equations instead of just one. This is confusing and begs the question of which equation we should take the derivative of. This is where implicit differentiation comes into play. Instead of making $y$ the subject, we keep the equation \eqref{eq:Fslf5tE3Tagbblgeb9s} as is.

We know from \eqref{eq:kRqwKaeXzuUpPstSdgE} that $y$ is a function of $x$. Therefore, we can replace $y$ with $y(x)$ to emphasize this fact. Keep in mind that $y(x)$ does not mean $y$ multiplied by $x$, but rather $y$ as a function of $x$. Let's now rewrite \eqref{eq:Fslf5tE3Tagbblgeb9s} as follows:

$$\begin{equation}\label{eq:RiPNMSOl5TT3Q7hlEou} x^2+[y(x)]^2=3^2 \end{equation}$$

Taking the derivative of both sides gives:

$$\begin{equation}\label{eq:EcLD65bmk58YD91kH3g} \frac{d}{dx} \Big(x^2+[y(x)]^2\Big) =\frac{d}{dx}(3^2) \end{equation}$$

The right-hand side is easy - the derivative of a constant is $0$ like so:

$$\frac{d}{dx}(3^2)=0$$

Let's now tackle the left-hand side:

\begin{equation}\label{eq:M7T97SIY3o8KSyK7fLW} \begin{aligned}[b] \frac{d}{dx} \Big(x^2+[y(x)]^2\Big)&= \frac{d}{dx}x^2+\frac{d}{dx}[y(x)]^2\\ &= 2x+\frac{d}{dx}[y(x)]^2\\ \end{aligned} \end{equation}

We must use the chain rule when taking the derivative of $[y(x)]^2$ with respect to $x$ because $y(x)$ is a function of $x$ like so:

$$\begin{equation}\label{eq:o2VbzIulAuTBKLY79o4} \frac{d}{dx}[y(x)]^2= 2\cdot{y(x)}\cdot{y'(x)} \end{equation}$$

Substituting \eqref{eq:o2VbzIulAuTBKLY79o4} into \eqref{eq:M7T97SIY3o8KSyK7fLW} gives:

$$\begin{equation}\label{eq:AhETI3i0iVFLLxA00HN} \frac{d}{dx} \Big(x^2+[y(x)]^2\Big)= 2x+2\cdot{y(x)}\cdot{y'(x)} \end{equation}$$

We now substitute \eqref{eq:AhETI3i0iVFLLxA00HN} back into our equation \eqref{eq:EcLD65bmk58YD91kH3g}:

$$2x+2\cdot{y(x)}\cdot{y'(x)}=0$$

Now that we are done with taking the derivative, we can drop the $(x)$ part:

$$2x+2\cdot{y}\cdot{y'}=0$$

Finally, let's make $y'$ the subject:

$$y'=-\frac{x}{y}$$

Unlike the standard case, the derivative of $y$ with respect to $x$ is dependent on both on $x$ and $y$, rather than just $x$.

Example.

# Computing slope of tangent line of a circle using implicit differentiation

Consider a circle with radius $5$:

$$x^2+y^2=5^2$$

Compute the slope of the tangent line at $(3,4)$.

Solution. Let's first confirm that the point $(3,4)$ is indeed on the circle:

$$(3)^2+(4)^2=25=5^2$$

Great, so the point $(3,4)$ lies on the circle! From the previous section, we know that the derivative of $y$ with respect to $x$ for a circle is computed by:

$$\begin{equation}\label{eq:A1sPwr15aeUpKbhW5mL} y'=\frac{dy}{dx}=-\frac{x}{y} \end{equation}$$

Let's substitute $x=3$ and $y=4$ into \eqref{eq:A1sPwr15aeUpKbhW5mL} to get:

$$\begin{equation}\label{eq:uxJaG0WbWBkywFiQI3G} y'=\frac{dy}{dx}=-\frac{3}{4} \end{equation}$$

This is the slope of the tangent line at $(3,4)$!

Finally, let's illustrate our findings: Great, we're done!

Example.

# Performing implicit differentiation

Consider the following function:

$$-2x^2+y=1$$

Compute the slope of the tangent line at point $(1,3)$.

Solution. We will tackle this problem in two ways:

• standard way of differentiation

• implicit differentiation

Unlike the circle example, we can easily make $y$ the subject to get the $f(x)$ form:

$$y=f(x)=2x^2+1$$

Taking the derivative of $f$ with respect to $x$ gives:

$$y'=4x$$

Therefore, at point $(1,3)$, the slope of the tangent line is:

$$y'=4$$

Let's now tackle this problem using implicit differentiation. We take the derivative of both sides with respect to $x$ like so:

\begin{align*} \frac{d}{dx}\Big(-2x^2+y(x)\Big)&=\frac{d}{dx}(1)\\ -4x+y'&=0\\ y'&=4x\\ \end{align*}

Therefore, at point $(1,3)$, the slope of the tangent line is:

$$y'=4$$

This is the same as the slope computed using the standard way of differentiation!

Finally, let's visualize the tangent line: Great, we're done!

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