Implicit Differentiation | Summary & Examples

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Summary

An implicit equation is an equation which is not in the form $y\quad =\quad f\left( x \right)$ , it consists of two variable x and y which cannot be separated.
Implicit Functions are differentiated by using ”chain rule” in combination with the ”product and quotient rule”.
When we differentiate y we write $\frac { dy }{ dx }$ with the derivative i.e
$\frac { d }{ dx } ({ y }^{ 2 })\quad =\quad 2y\frac { dy }{ dx }$
To find the derivative of the product of x and y i.e $\frac { d }{ dx } (xy)$ we use the product rule.

Previously, we have differentiated function where y was explicitly separated from x. Some examples of such functions $y\quad =\quad f\left( x \right)$ are:

$y\quad =\quad { x }^{ 2 }\quad +\quad 3x\quad +\quad 2$ $y\quad =\quad sin(x)$ $y\quad =\quad { e }^{ x }$

$y\quad =\quad { 5x\quad +\quad 1 }$ etc.

All these functions have y separated from x and we can then easily find derivative of y with respect to x $\frac { dy }{ dx }$ .

On the other hand, consider an equation ${ y }^{ 2 }\quad +\quad xy\quad +\quad y{ x }^{ 2 }\quad =\quad 2$ , we can see that we cannot separate y from x and we cannot write it in the form $y\quad =\quad f\left( x \right)$ , and therefore we cannot also find $\frac { dy }{ dx }$ directly differentiating the above equation. Such an equation is called and implicit function.

So to differentiate such an equation is known as ”Implicit Differentiation”.

Implicit Functions are differentiated by using ”chain rule” in combination with the product and quotient rule.

When we differentiate y we write $\frac { dy }{ dx }$ with the derivative i.e

$\frac { d }{ dx } ({ y }^{ 2 })\quad =\quad 2y\frac { dy }{ dx }$ $\frac { d }{ dx } ({ y }^{ 4 })\quad =\quad 4{ y }^{ 3 }\frac { dy }{ dx }$ $\frac { d }{ dx } ({ y }^{ 5 })\quad =\quad 5{ y }^{ 4 }\frac { dy }{ dx }$

Let’s move on to solve an example so we can apply these rules to differentiate an implicit function.

Example #1

Q. Differentiate ${ x }^{ 2 }\quad +\quad 3xy\quad +\quad { y }^{ 2 }\quad =\quad 0$

Solution:

We see that the above equation is implicit hence we differentiate each term with respect to x like:

${ x }^{ 2 }\quad +\quad 3xy\quad +\quad { y }^{ 2 }\quad =\quad 0$

$\frac { d }{ dx } ({ x }^{ 2 })\quad +\quad 3\frac { d }{ dx } (xy)\quad +\quad \frac { d }{ dx } ({ y }^{ 2 })\quad =\quad 0$

Use the product rule when differentiating (xy):

$2x\quad +\quad 3\left[ x\frac { dy }{ dx } \quad +\quad y\frac { dx }{ dx } \right] \quad +\quad 2y\frac { dy }{ dx } \quad =\quad 0$

$2x\quad +\quad 3x\frac { dy }{ dx } \quad +\quad 3y\quad +\quad 2y\frac { dy }{ dx } \quad =\quad 0$

$\frac { dy }{ dx } \left[ 3x\quad +\quad 2y \right] \quad +\quad 2x\quad +\quad 3y\quad =\quad 0$

Make $\frac { dy }{ dx }$ the subject of the equation:

$\frac { dy }{ dx } \quad =\quad \frac { -(2x\quad +\quad 3y) }{ 3x\quad +\quad 2y }$ Ans

Example #2

Q. Find the equation of the tangent to the curve ${ x }^{ 3 }\quad +\quad { y }^{ 3 }\quad =\quad 6xy$ at the point (3, 3)

Solution:

${ x }^{ 3 }\quad +\quad { y }^{ 3 }\quad =\quad 6xy$

$\frac { d }{ dx } ({ x }^{ 3 })\quad +\quad \frac { d }{ dx } ({ y }^{ 3 })\quad =\quad \frac { d }{ dx } (6xy)$

$3{ x }^{ 2 }\quad +\quad 3{ y }^{ 2 }\frac { dy }{ dx } \quad =\quad 6\left[ x\frac { dy }{ dx } \quad +\quad y\frac { dx }{ dx } \right]$

Move the terms with $\frac { dy }{ dx }$ together:

$3{ y }^{ 2 }\frac { dy }{ dx } \quad -\quad 6x\frac { dy }{ dx } \quad =\quad 6y\quad -\quad 3{ x }^{ 2 }$

$\frac { dy }{ dx } (3{ y }^{ 2 }\quad -\quad 6x)\quad =\quad 6y\quad -\quad 3{ x }^{ 2 }$

Make $\frac { dy }{ dx }$ the subject of the equation:

$\frac { dy }{ dx } \quad =\quad \frac { 6y\quad -\quad 3{ x }^{ 2 } }{ 3{ y }^{ 2 }\quad -\quad 6x }$

Simplify the expression:

$\frac { dy }{ dx } \quad =\quad \frac { 3(2y\quad -\quad { x }^{ 2 }) }{ 3({ y }^{ 2 }\quad -\quad 2x) }$

$\frac { dy }{ dx } \quad =\quad \frac { 2y\quad -\quad { x }^{ 2 } }{ { y }^{ 2 }\quad -\quad 2x }$

We can find the value of the gradient of the tangent by substituting x and y with the given point (3, 3):

$\frac { dy }{ dx } \quad =\quad \frac { 2(3)\quad -\quad { (3) }^{ 2 } }{ { (3) }^{ 2 }\quad -\quad 2(3) }$

$\frac { dy }{ dx } \quad =\quad \frac { -3 }{ 3 } \quad =\quad -1$

So the tangent line is the line with slope −1 through the point (3, 3).

Using the formula for equation of a line, we find the equation of the tangent to the curve at (3, 3):

$(y\quad -\quad { y }_{ 1 })\quad =\quad m(x\quad -\quad { x }_{ 1 })$

$m\quad =\quad -1$ and $({ x }_{ 1 },\quad { y }_{ 1 })\quad =\quad (3,\quad 3)$

$(y\quad -\quad 3)\quad =\quad -1(x\quad -\quad 3)$

$y\quad =\quad -x\quad +\quad 6$ Ans

Reference

https://npflueger.people.amherst.edu/math1a/lecture17.pdf