How do you implicitly differentiate #-y= x^3y^2-3x^2y^2-7x^2y^4 #?

Question

How do you implicitly differentiate #-y= x^3y^2-3x^2y^2-7x^2y^4    #?

Scarlett Allison · Accepted Answer

#dy/dx=(x^2y^2-6xy^2-14xy^4)/(28x^2y^3+6x^2y-2x^3y-1)#

This implicit differentiation makes heavy use of the product rule. Before differentiating the function, we should do a sample first:

#d/dx[5x^4y^2]=5y^2d/dx[x^4]+5x^4d/dx[y^2]=5y^2(4x^3)+5x^4(2y)dy/dx#

The most important part about this is that differentiating any term with a #y# will spit out a #dy/dx# term, thanks to the chain rule.

Differentiating the given function:

#-dy/dx=3x^2y^2+2x^3ydy/dx-6xy^2-6x^2ydy/dx-14xy^4-28x^2y^3dy/dx#

Solve for #dy/dx#.

#28x^2y^3dy/dx+6x^2ydy/dx-2x^3ydy/dx-dy/dx=3x^2y^2-6xy^2-14xy^4#

Factor out a #dy/dx# and divide:

#dy/dx=(x^2y^2-6xy^2-14xy^4)/(28x^2y^3+6x^2y-2x^3y-1)#

Emma Aldridge · Answer

undefined To implicitly differentiate the equation -y = x^3y^2 - 3x^2y^2 - 7x^2y^4, follow these steps:

1. Differentiate each term of the equation with respect to $x$.
2. Use the product rule and chain rule where necessary.
3. Solve for $\frac{dy}{dx}$ to find the derivative.

Applying these steps:

$$\frac{d}{dx}(-y) = \frac{d}{dx}(x^3y^2) - \frac{d}{dx}(3x^2y^2) - \frac{d}{dx}(7x^2y^4)$$

$$-\frac{dy}{dx} = 3x^2y^2 \frac{dy}{dx} + x^3(2y\frac{dy}{dx}) - 6xy^2 - 6x^2y\frac{dy}{dx} - 28xy^4 - 28x^2y^3\frac{dy}{dx}$$

Rearrange terms to solve for $\frac{dy}{dx}$:

$$=-6xy^2 - 28xy^4 + 3x^2y^2\frac{dy}{dx} + 2x^3y\frac{dy}{dx} - 6x^2y\frac{dy}{dx} - 28x^2y^3\frac{dy}{dx}$$

$$=-6xy^2 - 28xy^4 + (3x^2y^2 - 6x^2y - 28x^2y^3)\frac{dy}{dx} + 2x^3y\frac{dy}{dx}$$

$$=-6xy^2 - 28xy^4 + (3x^2y^2 - 6x^2y - 28x^2y^3 + 2x^3y)\frac{dy}{dx}$$

Finally, isolate $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{-6xy^2 - 28xy^4}{3x^2y^2 - 6x^2y - 28x^2y^3 + 2x^3y}$$