Question

Can you check if I did this right?
If 5x^2 + y^4 = −9 then evaluate the second derivative of y with respect to x when x = 2 and y = 1
So 5x^2 goes to 10x and y^4 goes to 4y^3 then that derivative is -10x/4y^3, then take the derivative of that using the quotient rule which gives you -40y^3 +120xy^2 / 16y^6, finally insert the given values and you get 12.5?

Answer 1

Did you evaluated the second derivative?

Answer 2

Not really. That's where I keep running into trouble. I know I'm doing something wrong, I just can't pinpoint it.

Answer 3

@mathmale @Zale101 Any idea or specifications to help?

Answer 4

\(5x^2 + y^4 = −9\)
First derivative
\(\frac{dy}{dx}(5x^2)+\frac{dy}{dx}(y^4)=\frac{dy}{dx}(-9)\)

\(10x+4y^3\frac{dy}{dx}=0\)

\(\frac{dy}{dx}=\frac{-10x}{4y^3}\)
\(\frac{dy}{dx}\)
Apply the second derivative:
\(\frac{d^2y}{dx^2}=\frac{-10(4y^3)+10x(12y^2\frac{dy}{dx})}{(4y^3)^2}\)

Now plug \(\frac{dy}{dx}=\frac{-10x}{4y^2}\) back to the second derivative

\(\frac{d^2y}{dx^2}=\frac{-10(4y^3)+10x(12y^2(\frac{-10x}{4y^3}))}{(4y^3)^2}\)


\(\frac{d^2y}{dx^2}=\frac{-10(4y^3)+10x(12y^2(\frac{-10x}{4y^3}))}{(4y^3)^2}\)

Now plug x=2 and y=1

\(y=-\frac{155}{2}\)

Answer 5

\(\frac{d^2y}{dx^2}=-\frac{155}{2}\)
Correction

Answer 6

So I forgot to plug back into the derivative dy/dx, which is why my answer was wonky? Thank you so much!!!

Answer 7

No problem!

Answer 8

Glad you've gotten help from Zale.
5x^2 + y^4 = −9 is an implicit function, by the way. You do not have to solve it for y before taking the derivative. Neither do you have to solve the result of the first differentiation for dy/dx before finding the 2nd derivative. If you want to explore this further, please let me (or Zale) know.

Answer 9

What do you mean? @mathmale @Zale101

Answer 10

He means if you need further help with implicit differentiation.
Thanks for showing your work when posting the problem!

Answer 11

Ohh, okie. Thank you for helping me! cx

Answer 12

My pleasure!