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I am trying to find the second derivative in this implicit differentiation problem. I can find the first derivative, but I am having trouble getting the second derivative.

First derivative:

$$\frac{d}{dx}(\sqrt{y}+4xy)= \frac12 y^{-1/2} (dy/dx)+(4y+4x(dy/dx))=0$$ $$dy/dx(1/(2\sqrt{y}) + 4x)=-4y$$ $$dy/dx= -(8y^{3/2})/(8\sqrt{y}x+1)$$

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  • $\begingroup$ Can you show your work up to now in finding the first derivative $\endgroup$ – Shailesh Oct 15 '15 at 3:21
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    $\begingroup$ I added my work $\endgroup$ – A. Cam Oct 15 '15 at 3:39
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$$dy/dx(1/(2\sqrt{y}) + 4x)=-4y$$
Differentiate both sides.
$$\frac{d^2y}{dx^2}(\frac{1}{2\sqrt{y}}+4x)+\frac{dy}{dx}(-\frac{1}{4y^{3/2}}\frac{dy}{dx}+4) = -4\frac{dy}{dx}$$
Then substitute the value of $\frac{dy}{dx}$.

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You have done most of the work. You properly obtained $$y'(x)=-\frac{8 y(x)^{3/2}}{8 x \sqrt{y(x)}+1}$$ Continue deriving $$y''(x)=\frac d {dx}\big(y'(x)\big)$$ and get, after simplifications, $$y''(x)=-\frac{4 \left(16 x y(x) y'(x)+3 \sqrt{y(x)} y'(x)-16 y(x)^2\right)}{\left(8 x \sqrt{y(x)}+1\right)^2}$$ Replacing $y'(x)$ by its expression leads, after still more simplifications, to $$y''(x)=\frac{32 \left(32 x \sqrt{y(x)}+5\right) y(x)^2}{\left(8 x \sqrt{y(x)}+1\right)^3}$$

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Just do it again.

From $y^{1/2}+4xy = 3 $, differentiating we get $0 = \frac12y^{-1/2}y'+4(y+xy') = y'(\frac12y^{-1/2}+4x)+4y $, or, as you got, $y' =\frac{-4y}{\frac12y^{-1/2}+4x} $.

To get $y''$, I would start with $0 =y'(\frac12y^{-1/2}+4x)+4y $. Differentiating this,

$\begin{array}\\ 0 &=y''(\frac12y^{-1/2}+4x)+y'(\frac12y^{-1/2}+4x)'+4y'\\ &=y''(\frac12y^{-1/2}+4x)+y'(-\frac14y^{-3/2}+4)+4y'\\ &=y''(\frac12y^{-1/2}+4x)+y'(-\frac14y^{-3/2}+8)\\ \end{array} $

so $y'' =\frac{y'(\frac14y^{-3/2}-8)}{\frac12y^{-1/2}+4x} $.

You can put $y'$ in there to get $y'' =\frac{\frac{-4y}{\frac12y^{-1/2}+4x}(\frac14y^{-3/2}-8)}{\frac12y^{-1/2}+4x} =\frac{-4y(\frac14y^{-3/2}-8)}{(\frac12y^{-1/2}+4x)^2} $.

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  • $\begingroup$ $(\frac12y^{-1/2}+4x)' = (-\frac14y^{-3/2}\color{red}{y'}+4)$ $\endgroup$ – Paul Sinclair Oct 31 '15 at 15:30

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