Implicit Differentiation

From Math Wiki
Revision as of 17:37, 23 November 2015 by MathAdmin (talk | contribs) (Created page with " == Background == So far, you may only have differentiated functions written in the form <math style="vertical-align: -5px">y=f(x)</math>. But some functions are better descr...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to navigation Jump to search


So far, you may only have differentiated functions written in the form . But some functions are better described by an equation involving and . For example, describes the graph of a circle with center and radius 4, and is really the graph of two functions , the upper and lower semicircles:

Upper semicircle.png Lower semicircle.png

Sometimes, functions described by equations in and are too hard to solve for , for example . This equation really describes 3 different functions of x, whose graph is the curve:


We want to find derivatives of these functions without having to solve for explicitly. We do this by implicit differentiation. The process is to take the derivative of both sides of the given equation with respect to , and then do some algebra steps to solve for (or if you prefer), keeping in mind that is a function of throughout the equation.

Warm-up exercises

Given that is a function of , find the derivative of the following functions with respect to .


Think , and view it as to see that the derivative is by the chain rule, but write it as .


and are both functions of which are being multiplied together, so the product rule says it's .


The function is inside of the cosine function, so the chain rule gives .


Write it as , and use the chain rule to get   , then simplify.

Exercise 1: Compute y'

Find if .

Note the term requires the chain rule, the   term needs the product rule, and the derivative of 8 is 0.

We get

Exercise 2: Find equation of tangent line

Find the equation of the tangent line to   at the point .

We first compute by implicit differentiation.

At the point , we have and . Plugging these into our equation for gives

This means the slope of the tangent line at is , and a point on this line is . Using the point-slope form of a line, we get

Here's a picture of the curve and tangent line:

Tangent line and curve.png

Exercise 3: Compute y"

Find if .

Use implicit differentiation to find first:

Now is just the derivative of with respect to . This will require the chain rule. Notice we already found the derivative of to be .


But we mustn't leave in our final answer. So, plug back in to get

as our final answer.

~Page created by Jordan Tousignant