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?
Did you evaluated the second derivative?
Not really. That's where I keep running into trouble. I know I'm doing something wrong, I just can't pinpoint it.
@mathmale @Zale101 Any idea or specifications to help?
\(5x^2 + y^4 = −9\) 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 \(y=-\frac{155}{2}\)
\(\frac{d^2y}{dx^2}=-\frac{155}{2}\) Correction
So I forgot to plug back into the derivative dy/dx, which is why my answer was wonky? Thank you so much!!!
No problem!
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.
What do you mean? @mathmale @Zale101
He means if you need further help with implicit differentiation. Thanks for showing your work when posting the problem!
Ohh, okie. Thank you for helping me! cx
My pleasure!
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