# Check my implicit differentiation please

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• Dec 6th 2010, 09:22 PM
Marconis
Check my implicit differentiation please
Question is,

Find an equation of tangent line:

x+1=(x+4y)/(2x+y^3) at (0,2).

Doing everything I got

y-2=-9/16 (x-0)

My steps to it took up half a page...I just can't type it all out. If my answer is incorrect, then I will so we can see where I went wrong. To me it was a hard and long problem.
• Dec 6th 2010, 10:42 PM
adkinsjr
I'm sure the problem is with the derivative.

$\displaystyle x+1=\frac{x+4y}{2x+y^3}$

$\displaystyle 1=\frac{(2x+y^3)'(x+4y)-(x+4y)'(2x+y^3)}{(2x+y^3)^2}$

$\displaystyle 1=\frac{(2+3y^2y')(x+4y)-(1+4y')(2x+y^3)}{(2x+y^3)^2}$

Can you take it from here? I'm sure your problem was finding the correct derivative y'.
• Dec 6th 2010, 10:46 PM
Glitch
Yeah, your derivative is wrong. You need to use implicit differentiation, then solve for $\displaystyle \frac{dy}{dx}$. I suggest multiplying by the denominator to avoid using the quotient rule.

I got $\displaystyle y = -\frac{9}{8}x + 2$
• Dec 7th 2010, 07:08 AM
Marconis
Ok, so I actually threw an extra y in by a mistake when finding the derviative of x+4y. Instead of it being 1+4y', I made it 1+4yy'. To solve after this point, do I multiply by (2x+y^3)^2 on both sides and then go from there?
• Dec 7th 2010, 07:10 AM
Marconis
Quote:

Originally Posted by Glitch
Yeah, your derivative is wrong. You need to use implicit differentiation, then solve for $\displaystyle \frac{dy}{dx}$. I suggest multiplying by the denominator to avoid using the quotient rule.

I got $\displaystyle y = -\frac{9}{8}x + 2$

I actually didn't know you were "allowed" to manipulate the problem like that to avoid using it. Cool! I'll do the problem later tonight after class and I'll let you know if I get the same answer. Thanks guys.