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.
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 . . .
5x^{2 }+ 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)
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 , and I implicitly differentiated y with respect to x to get this ... 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!
Should've posted it originally, my bad. Here it is:
(Took derivative with respect to x)
(multiplied 6 and -1 by 4)
(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.
(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?