Question

Find d^2y/dx^2 (second derivative) of x(siny) = 3

Answer 1

i found that dy/dx = -tany/x

Answer 2

so i got d^2y/dx^2 = -tany/x^2

Answer 3

is that right?

Answer 4

not sure with the second derivative

Answer 5

\[\frac{d^2y}{dx^2} = -\frac{ x\frac{dy}{dx}\sec^2 y - \tan y }{x^2}\]\[\frac{d^2y}{dx^2} = \frac{\tan y \sec^2 y + \tan y}{x^2}\]

Answer 6

ok thanks. lemme see what i did wrong.

Answer 7

wait. i think d^2y/dx^2 is wrong.
shouldn't it be

?
i can do the algebra from here...

Answer 8

you forget to multiply by -1, and dy/dx = -(tan y)/x, not -(tan y)/x^2

Answer 9

ohh i see. sorry i looked at the wrong dy/dx

Answer 10

Here's sort of what I'm getting \[
\begin{array}{rcl}
x(\sin y) &=& 3 \\
\frac{d}{dx}x(\sin y) &=& \frac{d}{dx}3 \\
\sin y + x\frac{dy}{dx}\cos y &=& 0 \\
x\frac{dy}{dx}\cos y &=& -\sin y \\
x\frac{dy}{dx} &=& -\tan y \\
\frac{dy}{dx} &=& \frac{ -\tan y}{x} \\
\frac{d}{dx} \left[ \frac{dy}{dx} \right] &=& \frac{d}{dx} \left[ \frac{- \tan y}{x} \right] \\
\frac{d^2y}{dx^2} &=& \frac{-\sec^2(y)\frac{dy}{dx}x-(-\tan y)}{x^2} \\
\end{array}
\]

Answer 11

Would have to sub in the \(\frac{-\tan y}{x}\) for \(\frac{dy}{dx}\) at the end.

Answer 12

yeah i know. thanks.
i think i originally took the wrong derivative