Question

If \frac(x^2)(64) + \frac (y^2)(25) = 1 and y( 2 ) = 4.84 , find y'( 2 ) by implicit differentiation.

Answer 1

Okay, so what do you get when you differentiate the first equation?

Answer 2

2x/64

Answer 3

It should still be an equation.

Answer 4

2x/64+(2yy')/25

Answer 5

=1

Answer 6

\[\frac d {dx} 1 = 1
\]Oh?

Answer 7

no it wasn't

Answer 8

the problem was changed on me so now its If \frac(x^2)(25) + \frac (y^2)(16) = 1 and y( 1 ) = 3.92 , find y'( 1 ) by implicit differentiation

Answer 9

Okay, so what is your equation after differentiating.

Answer 10

2x/25+2yy'/16=0

Answer 11

now, what is \(x\)?

Answer 12

You need to plug in the values of \(x\) and \(y\)

Answer 13

2/25+2(4.84)y'/16=0

Answer 14

Now solve for \(y'\).

Answer 15

I'm not going to get a medal, am I?

Answer 16

\( \dfrac{x^2}{25} + \dfrac {y^2}{16} = 1 \)

\( \dfrac{2x}{25} + \dfrac{2yy'}{16} = 0 \)

\( \dfrac{2(1)}{25} + \dfrac{2(3.92)y'}{16} = 0 \)

\( \dfrac{2(3.92)y'}{16} = -\dfrac{2}{25} \)

\( y' = -\dfrac{2(16)}{25(2)(3.92) } \)

\( y' = -\dfrac{8}{49 } \)