# Use implicit differentiation to find y' where x^2 +6xy + 4y^3 = 20. Evaluate this at (x, y) =...

## Question:

Use implicit differentiation to find y' where {eq}x^2 +6xy + 4y^3 = 20. {/eq}

Evaluate this at {eq}(x, y) = (2,1) {/eq}.

## Calculus

Calculus is one of the branches of mathematics. It deals with the study of change in the value of a function as the points in the domain change.

## Answer and Explanation: 1

**The value after differentiating at (x,y)=(2,1) is {eq}\frac{{ - 5}}{{12}}.
{/eq}**

Given: {eq}{x^2} + 6xy + 4{y^3} = 20 {/eq}

Differentiate it w.r.t. x,

{eq}\begin{align*} 2x\frac{{dx}}{{dx}} + 6x\frac{{dy}}{{dx}} + 6y\frac{{dx}}{{dx}} + 4\left( {3{y^2}} \right)\frac{{dy}}{{dx}} &= 0\\ 2x + 6xy' + 6y + 12{y^2}y' &= 0\\ 2\left( {x + 3xy' + 3y + 6{y^2}y'} \right) &= 0\\ x + 3xy' + 3y + 6{y^2}y' &= 0\\ x + 3y + y'\left( {3x + 6{y^2}} \right) &= 0\\ y'\left( {3x + 6{y^2}} \right) &= - \left( {x + 3y} \right)\\ y' &= \frac{{ - \left( {x + 3y} \right)}}{{3x + 6{y^2}}} \end{align*} {/eq}

The value of y' at (x,y)=(2,1) is:

{eq}\begin{align*} y' &= \frac{{ - \left( {2 + 3\left( 1 \right)} \right)}}{{3\left( 2 \right) + 6{{\left( 1 \right)}^2}}}\\ y' &= \frac{{ - \left( {2 + 3} \right)}}{{6 + 6}}\\ y' &= \frac{{ - 5}}{{12}} \end{align*} {/eq}

#### Learn more about this topic:

from

Chapter 6 / Lesson 5Explicit differentiation is used when 'y' is isolated, whereas implicit differentiation can be used similar to the chain rule when 'y' is not isolated. Learn more about implicit differentiation through examples of formulas and graphs.