Using Implicit Differentiation

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Math › Using Implicit Differentiation

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1

An ellipse is represented by the following equation:

$\small 8x^{2}+3y^{2}+6x+y-104 = 0$

What is the slope of the curve at the point (3,2)?

$\small \frac{-54}{13}$

$\small \frac{54}{13}$

$\small \frac{-13}{54}$

undefined

$\small 0$

Explanation

It would be difficult to differentiate this equation by isolating . Luckily, we don't have to. Use $\small \frac{dy}{dx}$ to represent the derivative of with respect to and follow the chain rule.

(Remember, $\small \frac{dx}{dx}$ is the derivative of with respect to , although it usually doesn't get written out because it is equal to 1. We'll write it out this time so you can see how implicit differentiation works.)

$\small 8x^{2}+3y^{2}+6x+y-104 = 0$

$\small 16x\frac{dx}{dx}+6y\frac{dy}{dx}+6\frac{dx}{dx}+\frac{dy}{dx}=0$

$\small 16x(1)+6y\frac{dy}{dx}+6(1)+\frac{dy}{dx}=0$

$\small 16x+6y\frac{dy}{dx}+6+\frac{dy}{dx}=0$

Now we need to isolate $\small \frac{dy}{dx}$ by first putting all of these terms on the same side:

$\small \small 6y\frac{dy}{dx}+\frac{dy}{dx}=-16x-6$

$\small \frac{dy}{dx}(6y+1)=-16x-6$

$\small \frac{dy}{dx}=\frac{-16x-6}{6y+1}$

This is the equation for the derivative at any point on the curve. By substituting in (3, 2) from the original question, we can find the slope at that particular point:

$\small \frac{dy}{dx}=\frac{-16(3)-6}{6(2)+1}=\frac{-48-6}{12+1}=\frac{-54}{13}$

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