1. 6.1 Implicit Vs Explicitap Calculus Calculator
2. 6.1 Implicit Vs Explicitap Calculus Algebra
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Finding the derivative when you can’t solve for y

6.1 Implicit Vs Explicitap Calculus Calculator

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

6.1worksheetans.pdf: File Size: 196 kb: File Type: Download File. Proudly powered by Weebly. Not all implicit equations can be restated explicitly in a single equation. For example, the implicit equation x 2 +y 2 = 9 needs two explicit equations, which are the top and bottom halves of a cricle respectively, to define the functional relation completely. Learning Objectives. 3.8.1 Find the derivative of a complicated function by using implicit differentiation.; 3.8.2 Use implicit differentiation to determine the equation of a tangent line. Please assume this is my initial exposure to calculus and derivatives. I am having difficulty making the connection between the application of the chain rule to explicit differentiation and that of implicit differentiation. Everything I’ve learned so far about differentiation has been based on explicitly defined functions and limits. The paper surveys different notions of implicit definition. In particular, we offer an examination of a kind of definition commonly used in formal axiomatics, which in general terms is understood as providing a definition of the primitive terminology of an axiomatic theory. We argue that such “structural definitions” can be semantically understood in two different ways, namely (1) as.

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:

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)

6.1 Implicit Vs Explicitap Calculus Algebra

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

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(f^½) = ddf(f^½)ddx(r² − x²)

Derivatives:

ddx(f^½) = 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.

6.1 Implicit Vs Explicitap Calculus Solver

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:

(the middle term is explained below)

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

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)

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

sin² y + cos² y = 1

cos y = √(1 − sin² y )

6.1 Implicit Vs Explicitap Calculus 2nd Edition

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

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

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