*Finding the derivative when you can’t solve for y *

- 6.1 Implicit Vs Explicitap Calculus Solver
- 6.1 Implicit Vs Explicitap Calculus Algebra
- 6.1 Implicit Vs Explicitap Calculus 2nd Edition

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.

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.*

Explicit Form | Implicit Form |

y = ± √ (r^{2} − x^{2}) | x^{2} + y^{2} = r^{2} |

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 x^{2} + y^{2} = 3^{2}

- Differentiate with respect to x
- Collect all the
*dy***dx**on one side - Solve for
*dy***dx**

Differentiate with respect to x:

*d***dx**(x^{2}) + *d***dx**(y^{2}) = *d***dx**(r^{2})

Let's solve each term:

Use the Chain Rule (explained below):*d***dx**(y^{2}) = 2y*dy***dx**

r^{2} is a constant, so its derivative is 0:*d***dx**(r^{2}) = 0

Which gives us:

2x + 2y*dy***dx** = 0

Collect all the *dy***dx** on one side

y*dy***dx** = −x

Solve for *dy***dx**:

*dy***dx** = *−x***y**

Let's look more closely at how *d***dx**(y^{2}) becomes 2y*dy***dx**

The Chain Rule says:

*du***dx** = *du***dy***dy***dx**

Substitute in u = y^{2}:

*d***dx**(y^{2}) = *d***dy**(y^{2})*dy***dx**

And then:

*d***dx**(y^{2}) = 2y*dy***dx**

Another common notation is to use ’ to mean *d***dx**

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) = y^{2}, so f’(y) = 2y:

f(y)’ = 2yy’

or alternatively: f(y)’ = 2y *dy***dx**

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

Square root:y = ±√(r^{2} − x^{2})

As a power: y = (r^{2} − x^{2})^{½}

Simplify:y’ = −x(r^{2} − x^{2})^{−½}

Now, because **y = (r**^{2} − x^{2})^{½}: y’ = −x/y

We get the same result this way!

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

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

*dy***dx** = *dy***df***df***dx**

Substitute in f = (r^{2} − x^{2}):

*d***dx**(f^{½}) = *d***df**(f^{½})*d***dx**(r^{2} − x^{2})

Derivatives:

*d***dx**(f^{½}) = ½(f^{−½}) (−2x)

And substitute back f = (r^{2} − x^{2}):

*d***dx**(r^{2} − x^{2})^{½} = ½((r^{2} − x^{2})^{−½}) (−2x)

And we simplified from there.

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

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

No problem, just substitute it into our equation:

*dy***dx** = −x/y

*dy***dx** = −3/4

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

y = −3/4 x + 25/4

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

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

- First, differentiate with respect to x (use the Product Rule for the xy
^{2}term). - Then move all dy/dx terms to the left side.
- Solve for dy/dx

Like this:

(the middle term is explained below)

Simplify :3(5y^{2}−6xy)*dy***dx**= 9y^{2} − 20x^{3}

And we get:

dydx = | 9y^{2} − 20x^{3} |

3(5y^{2} − 6xy) |

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

= x(2y*dy***dx**) + y^{2}

Because (y^{2})’ = 2y*dy***dx**(we worked that out in a previous example)

Oh, and *dx***dx** = 1, in other words x’ = 1

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:

In non−inverse mode:x = sin(y)

1 = cos(y) *dy***dx**

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

sin^{2} y + cos^{2} y = 1

cos y = √(1 − sin^{2} y )

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

cos y = √(1 − x^{2})

Which leads to:

*dy***dx**= *1***√(1 − x ^{2})**

So:y^{2} = x

Simplify:*dy***dx** = *1***2y**

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

As a power:y = x^{½}

Simplify:*dy***dx** = *1***2√x**

- 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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