Question

If (x^2/16) + (y^2/49) = 1 and y(2)=(7/4)sqrt12 Find y'(2) by implicit differentiation

Answer 1

Please help I am very confused and lost as to what to do with this problem.

Answer 2

Do you know how to find y' ?

Answer 3

I don't know I don't know if this is correct but this is what I started doing I tried to find the derivative of the one with the x and y's and got (32x/16^2)+ (98y/49^2)(dy/dx)

Answer 4

Hmm..
\[\frac{x^2}{16} + \frac{y^2}{49} = 1\]Diff. both sides w.r.t. x.
How do you differentiate the first term on the left?

Answer 5

I tried to do the bottom multiplied by the derivative of the top - top multiplied by the derivative of the bottom over the bottom squared for the left and then the derivative of 1 just goes to 0

Answer 6

Hmm..
So, we agree that derivative of one is zero. No problem.
The problem is finding the derivative of the term on the left.
\[\frac{d}{dx}(\frac{x^2}{16} + \frac{y^2}{49})=\frac{d}{dx}(\frac{x^2}{16}) + \frac{d}{dx}(\frac{y^2}{49})\]So far so good?

Answer 7

Yes

Answer 8

Nice~
Consider the first one
\[\frac{d}{dx}(\frac{x^2}{16}) = \frac{1}{16}(\frac{d}{dx}x^2)\]Happy with this?

Answer 9

Okay so then y would be 1/49((d/dx)y^2) correct?

Answer 10

Oh yes!
Can you find the derivatives again?

Answer 11

So do you mean by the inside which would be d/dx 2x and d/dx 2y?

Answer 12

Hmm, not really correct.
What is the derivative of x^2 with respect to x?

Answer 13

2x

Answer 14

Yes. So, \(\frac{1}{16}(\frac{d}{dx}x^2) = \frac{1}{16}(2x)\), right?

Answer 15

yes so then 1/49(2y) is the other part correct?

Answer 16

1/49(2y) => not correct.

Answer 17

Imagine this..
\[\frac{d}{dx}x^2 = 2x\frac{dx}{dx} = 2x\]\[\frac{d}{dx}y^2 = ...?\]

Answer 18

2y then would follow that logic correct?

Answer 19

Nope, not really, instead, it should be (2y)y'