Question

Could someone help me take the second derivative using implicit differentiation?

Answer 1

I have a question and i did it, but im stuck on a part

Answer 2

the first deri is -x^2/y^2

Answer 3

Just to be sure we are on the right track, what is the original equation and question?

Answer 4

x^3 +y^3 =16....find second deriv at (2,2)

Answer 5

taking the derivative with respect to x. I do
\[ \frac{d}{dx} \left( x^3 +y^3 =16\right)\\
\frac{d}{dx}x^3 + \frac{d}{dx}y^3 = \frac{d}{dx}16\\
3x^2 \frac{dx}{dx}+3y^2 \frac{dy}{dx}=0\\
3y^2 \frac{dy}{dx}= -3x^2\\
\frac{dy}{dx}= x^2 y^-2
\]

Answer 6

so you did the first derivative correctly. now on to the second.
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(x^2y^{-2} ) \\
= x^2 \frac{d}{dx}y^{-2} + y{-2}\frac{d}{dx}x^2
\]
can you do the next steps

Answer 7

can i show u what i did

Answer 8

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(x^2y^{-2} ) \\
= x^2 \frac{d}{dx}y^{-2} + y^{-2}\frac{d}{dx}x^2 \]

Answer 9

\[= \left(\begin{matrix}[-2x](y^2) - [2y dy/dex](-x^2) \\ y^4\end{matrix}\right)\]

Answer 10

thats what i have
i mess up when i sub in the first derive

Answer 11

are you using the quotient rule ?

Answer 12

yes

Answer 13

I think the signs are wrong. Try writing it as x^2 y^(-2) and use the product rule.
we can come back to the quotient rule after we get the right answer.

Answer 14

ok

Answer 15

omg omg i got the right answer. thank you sooooo much

Answer 16

for the quotient rule you should do
\[\frac{d}{dx}\left( \frac{x^2}{y^2}\right)= \frac{y^2 \frac{d}{dx}x^2 - x^2 \frac{d}{dx}y^2}{y^4}\]

Answer 17

the numerator gives you
\[ y^2 \frac{d}{dx}x^2 - x^2 \frac{d}{dx}y^2 \\
y^2 (2x) - x^2 2y y' \\
2x y^2 - 2x^2 y \frac{x^2}{y^2}\\
2xy^2 - 2\frac{x^4}{y}
\]
now divide all that by y^4
\[ \frac{1}{y^4} \left( 2xy^2 - 2\frac{x^4}{y}\right)\\
\frac{2x}{y^2} - \frac{2x^4}{y^5}
\]