How do you implicitly differentiate #x^4+2y^2=8 #?

Question

Evelyn Agard · Accepted Answer

#(dy)/(dx)= -x^3/y# #(dx^4)/(dx) = 4x^3# #(d2y^2)/(dx) = (d2y^2)/(dy)*(dy)/(dx) = 4y(dy)/(dx)# #(d (x^4+2y^2))/(dx) = (dx^4)/(dx)+(d2y^2)/(dx) = 4x^3+4y(dy)/(dx)# #(d8)/(dx) = 0# Since #x^4+2y^2=8# #color(white)("XXX")(d(x^4+2y^2))/(dx) = (d8)/(dx)# #color(white)("XXX")4x^3+4y(dy)/(dx)=0# #color(white)("XXX")4y(dy)/(dx) = -4x^3# #color(white)("XXX")(dy)/(dx) = (-4x^3)/(4y)=-x^3/y#

Adam Adkins · Answer

undefined To implicitly differentiate the equation $x^4 + 2y^2 = 8$:

1. Differentiate each term of the equation with respect to $x$.
2. Treat $y$ as a function of $x$ and apply the chain rule whenever differentiating $y$.

Differentiating $x^4$ with respect to $x$ yields $4x^3$.
Differentiating $2y^2$ with respect to $x$ yields $4y \frac{dy}{dx}$ by the chain rule.
The derivative of a constant (in this case, 8) is 0.

So, the implicit differentiation of $x^4 + 2y^2 = 8$ is $4x^3 + 4y \frac{dy}{dx} = 0$.