# Beginner Implicit differentiation

• Mar 8th 2013, 12:08 PM
jjtjp
Beginner Implicit differentiation
Here's the problem:
$x^4+x^2y+8y^2=10$, Find y' by implicit differentiation.

I set it up like this by taking the derrivative of both sides:
$\frac{\mathrm{d} y}{\mathrm{d} x}[x^4]+\frac{\mathrm{d} y}{\mathrm{d} x}[x^2y]+\frac{\mathrm{d} y}{\mathrm{d} x}[8y^2]=\frac{\mathrm{d} y}{\mathrm{d} x}[10]$

Giving me:
$4x^3+2xy+x^2y'+16yy'=0$

then:
$x^2y'+16yy'=4x^3+2xy$

Factor out y':
$y'(x^2+16y)=4x^3+2xy$

$y'=\frac{4x^3+2xy}{x^2+16y}$

This answer is wrong though. Can someone help me see where I went wrong? I'm wondering if ${\frac{\mathrm{d} y}{\mathrm{d} x}}{[x^2y]}\neq (2xy+x^2y')$

• Mar 8th 2013, 12:32 PM
a tutor
Re: Beginner Implicit differentiation
Quote:

Originally Posted by jjtjp
Here's the problem:
$x^4+x^2y+8y^2=10$, Find y' by implicit differentiation.

I set it up like this by taking the derrivative of both sides:
$\frac{\mathrm{d} y}{\mathrm{d} x}[x^4]+\frac{\mathrm{d} y}{\mathrm{d} x}[x^2y]+\frac{\mathrm{d} y}{\mathrm{d} x}[8y^2]=\frac{\mathrm{d} y}{\mathrm{d} x}[10]$

Giving me:
$4x^3+2xy+x^2y'+16yy'=0$

then:
$x^2y'+16yy'=4x^3+2xy$ This should be $x^2y'+16yy'=-4x^3-2xy$

Factor out y':
$y'(x^2+16y)=4x^3+2xy$

$y'=\frac{4x^3+2xy}{x^2+16y}$

This answer is wrong though. Can someone help me see where I went wrong? I'm wondering if ${\frac{\mathrm{d} y}{\mathrm{d} x}}{[x^2y]}\neq (2xy+x^2y')$

See above.
• Mar 8th 2013, 12:44 PM
jjtjp
Re: Beginner Implicit differentiation
Wow..I solved the problem 3 times and missed it every time. How the heck?! Anyways, thank you!
• Mar 8th 2013, 01:19 PM
HallsofIvy
Re: Beginner Implicit differentiation
Quote:

Originally Posted by jjtjp
Here's the problem:
$x^4+x^2y+8y^2=10$, Find y' by implicit differentiation.

I set it up like this by taking the derrivative of both sides:
$\frac{\mathrm{d} y}{\mathrm{d} x}[x^4]+\frac{\mathrm{d} y}{\mathrm{d} x}[x^2y]+\frac{\mathrm{d} y}{\mathrm{d} x}[8y^2]=\frac{\mathrm{d} y}{\mathrm{d} x}[10]$

Do NOT write " $\frac{\mathrm{d} y}{\mathrm{d} x}[x^4]$", the "y" should not be there. You mean $\frac{\mathrm{d} x^4}{\mathrm{d} x}$ or $\frac{\mathrm{d} }{\mathrm{d} x}[x^4]$

Quote:

Giving me:
$4x^3+2xy+x^2y'+16yy'=0$

then: $x^2y'+16yy'=4x^3+2xy$
NO, you have subtracted $4x^3+ 2xy$ from both sides. This should be $x^2y'+ 16yy'= -(4x^3+ 2xy)$ or $x^2y'+ 16yy'= -4x^3- 2xy$

Quote:

Factor out y':
$y'(x^2+16y)=4x^3+2xy$

$y'=\frac{4x^3+2xy}{x^2+16y}$
This answer is wrong though. Can someone help me see where I went wrong? I'm wondering if ${\frac{\mathrm{d} y}{\mathrm{d} x}}{[x^2y]}\neq (2xy+x^2y')$