Solving x^2+y^2=25: Finding the Value of d^2y/dx^2 at (4,3)

[UrbanXrisis]

if $$x^2+y^2=25$$, what is the value of $$\frac{d^2y}{dx^2}$$ at the point (4,3)?

$$x^2+y^2=25$$
$$2x+2y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=-\frac{2x}{2y}$$
$$\frac{dy}{dx}=-\frac{3}{4}$$
$$\frac{d^2y}{dx^2}=\frac{-y+\frac{dy}{dx}(-x)}{x^2}$$
$$\frac{d^2y}{dx^2}=\frac{-3+-\frac{3}{4}(-4)}{4^2}$$
$$\frac{d^2y}{dx^2}=0$$

where did I go wrong?

 

[jamesrc]

When you computed the 2nd derivative, I assume you used the quotient rule, but you have a typo in your denominator and you missed a sign in your second term.

 

[UrbanXrisis]

here's what I have $$\frac{d^2y}{dx^2}=\frac{-y+\frac{dy}{dx}(x)}{y^2}$$

is that correct?

 

[UrbanXrisis]

$$x^2+y^2=25$$
$$2x+2y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=-\frac{x}{y}$$
$$\frac{dy}{dx}=-\frac{4}{3}$$
$$\frac{d^2y}{dx^2}=\frac{-y+\frac{dy}{dx}(x)}{y^2}$$
$$\frac{d^2y}{dx^2}=\frac{-3+-\frac{4}{3}(4)}{3^2}$$
$$\frac{d^2y}{dx^2}=-\frac{25}{27}$$

is that correct?

 

[dextercioby]

Nope,the differentiation is okay,but the arithmetics is terrible...

Daniel.

 

[UrbanXrisis]

whoops forgot to change my previous numbers that I just copied, it's edited

 

[dextercioby]

It looks okay,now.

Daniel.