10/8/2021

## 6.1 Implicit Vs Explicitap Calculus

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Finding the derivative when you can’t solve for y

Implicit Differentiation In many examples, especially the ones derived from differential equations, the variables involved are not linked to each other in an explicit way. Most of the time, they are linked through an implicit formula, like F ( x, y ) =0. Every y=f (x) is an explicit function because it is clear that the value of y is dependent on the value of x. On the other side, an implicit function is any 'function' where there doesn't appear to be any dependent variable, such as x^2+y^2=1. To put simply because the calculators were designed as such. Lets have some history. First Casio calculators made before 1978 or so had no operator precedence. They calculated everything left to right so 1+2.3 gave 9.

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 Form Implicit 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

## 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

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

y = −3/4 x + 25/4

## Another Example

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!

• 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:

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

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.

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:

## 6.1 Implicit Vs Explicitap Calculus Solver

sin2 y + cos2 y = 1 cos y = √(1 − sin2 y )

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

cos y = √(1 − x2) dydx= 1√(1 − x2)

## 6.1 Implicit Vs Explicitap Calculus 2nd Edition

### 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

## 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