How do you implicitly differentiate #-1=y^3x-xy-9x^4y #?

Question

How do you implicitly differentiate #-1=y^3x-xy-9x^4y   #?

Christopher Agresta · Accepted Answer

You can do it like this:

#-1=y^3x-xy-9x^4# #D(-1)=D(y^3x-xy-9x^4y)# #0=y^3cancel(d(x))/(cancel(dx))+x.3y^2.y'-(xy'+ycancel(d(x))/(cancel(dx)))-9(x^4y'+y.36x^3)# #0=y^3+x.3y^2.y'-xy'-y-9x^4y'-9y.36x^3# #:.0=y^3+y'(x3y^2-x-9x^4)-9y.36x^3# #:.y'=((324x^3-y^3))/(x(3y^2-1-9x^3))#

Adam Adkins · Answer

undefined To implicitly differentiate the given equation, $ -1 = y^3x - xy - 9x^4y $, with respect to $x$, follow these steps:

1. Differentiate each term of the equation with respect to $x$.
2. Use the chain rule for terms involving $y$ since $y$ is a function of $x$.
3. Simplify the resulting expression.

Differentiating each term:

$$\frac{d}{dx}(-1) = 0$$

$$\frac{d}{dx}(y^3x) = y^3 + 3y^2 \cdot \frac{dy}{dx}$$

$$\frac{d}{dx}(-xy) = -y - x\frac{dy}{dx}$$

$$\frac{d}{dx}(-9x^4y) = -36x^3y - 9x^4\frac{dy}{dx}$$

Combine the terms:

$$0 = y^3 + 3y^2 \cdot \frac{dy}{dx} - y - x\frac{dy}{dx} - 36x^3y - 9x^4\frac{dy}{dx}$$

Now, solve for $\frac{dy}{dx}$:

$$0 = (3y^2 - 1 - 9x^4)\frac{dy}{dx} - y - 36x^3y$$

$$3y^2 - 1 - 9x^4 \neq 0$$

So, the implicit derivative is:

$$\frac{dy}{dx} = \frac{y + 36x^3y}{3y^2 - 1 - 9x^4}$$