# I need to find $\frac{d^{2}y}{dx^{2}}$ of $\sqrt{y} + 4xy = 4$

So I'm able to find the first derivative which comes out to be:

$$\frac{-8y^{3/2}}{8x\sqrt{y} +1}$$

From there I have to do the quotient rule with $-8y^{3/2}$ as my $f(x)$ and $8x\sqrt{y}+1$ as my $g(x)$. For $f'(x)$ I do -8 * ${3\over2}$ * $\sqrt{y}$ * $y'$ with $y'$ being = $\frac{-8y^{3/2}}{8x\sqrt{y} +1}$

This brings me to $f'(x)$ = $\displaystyle\frac{96y^2}{8x\sqrt{y}+1}$.

For $g'(x)$ I do 8 * the product rule of $x$ and $\sqrt{y}$

8((1*$\sqrt{y}$) + ($x$ * ${1\over2}$ * $y^{-1/2}$ * $y'$)) again with $y'$ = $\frac{-8y^{3/2}}{8x\sqrt{y} +1}$

Now for my $g'(x)$ I have a problem. I'm not sure if it would be $\displaystyle \frac{32xy + 8\sqrt{y}}{8x\sqrt{y}+1}$ OR just $\displaystyle \frac{32}{8x\sqrt{y}+1}$.

• State with respect to what variable you're differentiating. Commented Oct 5, 2014 at 21:34
• This is with respect to x Commented Oct 5, 2014 at 21:46
• I am not sure if I understand your approach, but I have given an answer below. Actually, the quotient rule works fine, too. Commented Oct 5, 2014 at 22:05
• Sorry, but you edited again your title... What's going to mean now? What about the equality? My answer below just concerns the evaluation of the second-order derivative of $f$. Commented Oct 5, 2014 at 22:41

Let $$f(x,y)=\frac{-8y^{3/2}}{8x\sqrt{y}+1},$$ then $$\frac{\partial f}{\partial x}= \Big(-8y^{3/2}\Big)\Bigg(-\frac{1}{\big(8x\sqrt{y}+1\big)^2}\Bigg)\frac{\partial}{\partial x}\Big(8x\sqrt{y}+1\Big),$$ which leads to $$\frac{\partial f}{\partial x}= \frac{8y^{3/2}(8\sqrt{y}+1)}{\big(8x\sqrt{y}+1\big)^2}.$$ Now, you need to take the following and do some mathematical manipulations, $$\frac{\partial^2}{\partial x^2}= \frac{\partial}{\partial x}\Big(\frac{\partial f}{\partial x}\Big).$$

• So, you need to find the second-order derivative now? You edited your title. Ok, follow the above rule and you'll find out the final answer. Commented Oct 5, 2014 at 22:27
• So it told me the answer is suppose to be $$\frac{32y^{2}(5+32x\sqrt{y})}{(1+8x\sqrt{y})^{3}}$$ Sorry I guess I typed it wrong. I was trying to follow what the book shows. Commented Oct 5, 2014 at 22:29