Post

Created by @nathanedwards
 at October 31st 2023, 8:17:03 am.

Implicit Differentiation Exam Question

Consider the equation:

x3+y33xy+1=0 x^3 + y^3 - 3xy + 1 = 0
  1. Use implicit differentiation to find dydx\frac{{dy}}{{dx}} in terms of xx and yy.

Solution:

To find dydx\frac{{dy}}{{dx}} using implicit differentiation, we differentiate both sides of the given equation with respect to xx.

ddx(x3+y33xy+1)=ddx(0) \frac{{d}}{{dx}}(x^3 + y^3 - 3xy + 1) = \frac{{d}}{{dx}}(0)

Differentiating each term separately, we get:

ddx(x3)+ddx(y3)ddx(3xy)+ddx(1)=0 \frac{{d}}{{dx}}(x^3) + \frac{{d}}{{dx}}(y^3) - \frac{{d}}{{dx}}(3xy) + \frac{{d}}{{dx}}(1) = 0

Using the power rule, chain rule, and product rule, the derivatives are:

3x2+3y2dydx3(xdydx+y)+0=0 3x^2 + 3y^2 \frac{{dy}}{{dx}} - 3\left(x\frac{{dy}}{{dx}} + y\right) + 0 = 0

Now, we isolate dydx\frac{{dy}}{{dx}} term:

3y2dydx3xdydx=3x2+3y 3y^2 \frac{{dy}}{{dx}} - 3x\frac{{dy}}{{dx}} = -3x^2 + 3y

Factor out dydx\frac{{dy}}{{dx}}:

dydx(3y23x)=3x2+3y \frac{{dy}}{{dx}}(3y^2 - 3x) = -3x^2 + 3y

Finally, divide both sides by 3y23x3y^2 - 3x to solve for dydx\frac{{dy}}{{dx}}:

dydx=3x2+3y3y23x \frac{{dy}}{{dx}} = \frac{{-3x^2 + 3y}}{{3y^2 - 3x}}

Hence, dydx\frac{{dy}}{{dx}} in terms of xx and yy is:

dydx=3x2+3y3y23x \frac{{dy}}{{dx}} = \frac{{-3x^2 + 3y}}{{3y^2 - 3x}}

Therefore, the answer is: [ \frac{{dy}}{{dx}} = \frac{{-3x^2 + 3y}}{{3y^2 - 3x}} ]