4.1 Implicit Differentiationap Calculus



Finding the derivative when you can’t solve for y

Showing 17 items from page AP Calculus Differentiation Homework sorted by Assignment Number. View more » Extra Practice. Assignment Answer Key. Implicit differentiation is a way of differentiating when you have a function in terms of both x and y. For example: x2 +y2 = 16. This is the formula for a circle with a centre at (0,0) and a radius of 4. So using normal differentiation rules x2 and 16 are differentiable if we are differentiating with respect to x. This lesson contains the following Essential Knowledge (EK) concepts for the.AP Calculus course.Click here for an overview of all the EK's in this course. EK 2.1C5. AP® is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site.

You may like to read Introduction to Derivatives and Derivative Rules first.

Implicit vs Explicit

A function can be explicit or implicit:

Explicit: 'y = some function of x'. When we know x we can calculate y directly.

Implicit: 'some function of y and x equals something else'. Knowing x does not lead directly to y.

Example: A Circle

Explicit FormImplicit Form
y = ± √ (r2 − x2)x2 + y2 = r2
In this form, y is expressed
as a function of x.		In this form, the function is
expressed in terms of both y and x.


The graph of x2 + y2 = 32

How to do Implicit Differentiation

  • Differentiate with respect to x
  • Collect all the dydx on one side
  • Solve for dydx

Example: x2 + y2 = r2

Differentiate with respect to x:

ddx(x2) + ddx(y2) = ddx(r2)

Let's solve each term:

Use the Chain Rule (explained below):ddx(y2) = 2ydydx
r2 is a constant, so its derivative is 0:ddx(r2) = 0

Which gives us:

2x + 2ydydx = 0

Collect all the dydx on one side

ydydx = −x

Solve for dydx:

dydx = −xy

The Chain Rule Using dydx

Let's look more closely at how ddx(y2) becomes 2ydydx

The Chain Rule says:

dudx = dudydydx

Substitute in u = y2:

ddx(y2) = ddy(y2)dydx

And then:

ddx(y2) = 2ydydx

Basically, all we did was differentiate with respect to y and multiply by dydx

Another common notation is to use ’ to mean ddx

The Chain Rule Using ’

The Chain Rule can also be written using ’ notation:

f(g(x))’ = f’(g(x))g’(x)

g(x) is our function 'y', so:

f(y)’ = f’(y)y’

f(y) = y2, so f’(y) = 2y:

f(y)’ = 2yy’

or alternatively: f(y)’ = 2y dydx

Again, all we did was differentiate with respect to y and multiply by dydx

Explicit

Let's also find the derivative using the explicit form of the equation.

  • To solve this explicitly, we can solve the equation for y
  • Then differentiate
  • Then substitute the equation for y again

Example: x2 + y2 = r2

Square root:y = ±√(r2 − x2)
As a power: y = (r2 − x2)½
Simplify:y’ = −x(r2 − x2)−½
Now, because y = (r2 − x2)½: y’ = −x/y

We get the same result this way!

You can try taking the derivative of the negative term yourself.

Chain Rule Again!

Yes, we used the Chain Rule again. Like this (note different letters, but same rule):

dydx = dydfdfdx

Substitute in f = (r2 − x2):

ddx(f½) = ddf(f½)ddx(r2 − x2)

Derivatives:

ddx(f½) = ½(f−½) (−2x)

And substitute back f = (r2 − x2):

ddx(r2 − x2)½ = ½((r2 − x2)−½) (−2x)

And we simplified from there.

Using The Derivative

OK, so why find the derivative y’ = −x/y ?

Well, for example, we can find the slope of a tangent line.

Example: what is the slope of a circle centered at the origin with a radius of 5 at the point (3,4)?

No problem, just substitute it into our equation:

dydx = −x/y

dydx = −3/4

4.1 implicit differentiationap calculus solver

And for bonus, the equation for the tangent line is:

y = −3/4 x + 25/4

Another Example

4.1 Implicit Differentiationap Calculus Solver

Sometimes the implicit way works where the explicit way is hard or impossible.

Example: 10x4 - 18xy2 + 10y3 = 48

How do we solve for y? We don't have to!

4.1 Implicit Differentiationap Calculus
  • First, differentiate with respect to x (use the Product Rule for the xy2 term).
  • Then move all dy/dx terms to the left side.
  • Solve for dy/dx

Like this:

4.1 Implicit Differentiationap Calculus
Derivative:10 (4x3) − 18(x(2ydydx) + y2) + 10(3y2dydx) = 0

(the middle term is explained below)

dydx on left:−36xydydx + 30y2dydx = −40x3 + 18y2
Simplify :3(5y2−6xy)dydx= 9y2 − 20x3

And we get:

dydx = 9y2 − 20x3
3(5y2 − 6xy)

Product Rule

For the middle term we used the Product Rule: (fg)’ = f g’ + f’ g

= x(2ydydx) + y2

4.1 Implicit Differentiationap Calculus Calculator

Because (y2)’ = 2ydydx(we worked that out in a previous example)

Oh, and dxdx = 1, in other words x’ = 1

Inverse Functions

Implicit differentiation can help us solve inverse functions.

The general pattern is:

  • Start with the inverse equation in explicit form. Example: y = sin−1(x)
  • Rewrite it in non-inverse mode: Example: x = sin(y)
  • Differentiate this function with respect to x on both sides.
  • Solve for dy/dx

As a final step we can try to simplify more by substituting the original equation.

4.1 Implicit Differentiationap Calculus Multiple Choice

An example will help:

Example: the inverse sine function y = sin−1(x)

In non−inverse mode:x = sin(y)
1 = cos(y) dydx

We can also go one step further using the Pythagorean identity:

sin2 y + cos2 y = 1

cos y = √(1 − sin2 y )

And, because sin(y) = x (from above!), we get:

cos y = √(1 − x2)

Which leads to:

dydx= 1√(1 − x2)

Example: the derivative of square root √x

So:y2 = x
Simplify:dydx = 12y

Note: this is the same answer we get using the Power Rule:

As a power:y = x½
Simplify:dydx = 12√x

4.1 Implicit Differentiationap Calculus Algebra

Summary

  • To Implicitly derive a function (useful when a function can't easily be solved for y)
    • Differentiate with respect to x
    • Collect all the dy/dx on one side
    • Solve for dy/dx
  • To derive an inverse function, restate it without the inverse then use Implicit differentiation




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