Implicit Differentiation

**phenol** · October 1st 2012, 08:17 PM

Hello,
Would anyone be able to tell me how to go about solving this problem? I've been looking through my calculus (it doesn't give examples similar to this problem) book but I have no idea where to start. Any help would be greatly appreciated. For the 1st part, I got dy/dx = -(10x+3+y)/x. Not sure if it's correct or not. Am I supposed to plug 2 in to find the slope? Thanks! Ana

Here's the problem . . .
5x²+ 3x + xy = 2 and y(2) = -12, find y'(2) by implicit differentiation. (1st part)

2nd part asks to find the equation of the tangent line to the graph at the point (2,-12)

**MarkFL** · October 1st 2012, 08:38 PM

You differentiated correctly, now use the given point (2,-12) to find y'(2). Then use the point-slope formula to find the equation of the tangent line.

**AZach** · October 1st 2012, 09:16 PM

Instead of making a new thread on these, I want to just pop in a quick question to see if I implicitly differentiated this correctly; problem is $(6x-y)^4+3y^3=9976$ , and I implicitly differentiated y with respect to x to get this $dy/dx = \frac{(24(6x-y)^3)}{(1+9y^2)}$ ... Does that look about right? I have a gut feeling it isn't, but I've combed through it a few times and I'm not sure how else to differentiate it. Thanks!

**MarkFL** · October 1st 2012, 09:48 PM

I get a different result...can you post your work?

**phenol** · October 1st 2012, 09:53 PM

Thanks for your help, I think I got it. The slope came out to -11/2 and the equation for the tangent line came out to be y=-11/2(x)-1. Thanks again!!

**MarkFL** · October 1st 2012, 09:57 PM

Yes, I get the same result.

**AZach** · October 2nd 2012, 11:03 AM

Originally Posted by MarkFL2

I get a different result...can you post your work?

Should've posted it originally, my bad. Here it is:

$(6x-y)^4+3y^3=9976$

$d/dx[(6x-y)^4]+d/dx[3y^3]=d/dx[9976]$ (Took derivative with respect to x)

$4(6x-y)^3*6*(-1)*(-dy/dx)+9y^2*dy/dx=0$

$-24(6x-y)^3*(-dy/dx)+9y^2*dy/dx=0$ (multiplied 6 and -1 by 4)

$dy/dx(24(6x-y)^3+9y^2)=0$ (factored out a dy/dx and got rid of the -1 by multiplying it by -24)

This is where I'm confused, and I have a feeling it's because I'm not doing something right when I differentiated the inside function -y from (6x-y)^4?

Or I should revisit step 4 and divide the left side and right side of the equation by -24.

$\frac{-24(6x-y)^3*(-dy/dx)+9y^2*dy/dx}{-24}=\frac{0}{-24}$

$dy/dx[(6x-y)^3*-1+9y^2]=0$ (now factor out dy/dx)

Actually that's still going to be wrong because dy/dx isn't going to be zero. I must need to add or subtract something from the left to get a quantity on the right?

Thread: Implicit Differentiation

LinkBack

Thread Tools

Display