# How do you get the second order implicit differentiation?

We were given an exercise in school:

Given $x^2 + 25y^2 = 100$, show that $$\frac{dy}{dx^2} = -\frac{4}{25y^3}$$

I am stuck on the first order which is $$-\frac{x}{25y}$$ When I'm now going to the second, I always end up with the wrong answer where the denominator is somewhere at $625y^2$. I know that you have to use the quotient rule on the first order derivative to get to the second, but I don't know how. Can anyone answer with a detailed step-by-step solution? I'm really lost. Our prof didn't explain that topic well to us.

• Please see this tutorial on how to format mathematics on this site. – N. F. Taussig Mar 22 '15 at 12:35

hint: $(x^2+25y^2)' = 0 \to 2x+50yy' = 0 \to y' = -\dfrac{x}{25y} \to y'' = \left(\dfrac{-x}{25y}\right)' = \dfrac{....}{625y^2}$
You were given $$x^2 + 25y^2 = 100\tag{1}$$ from which you correctly determined that $$y' = -\frac{x}{25y}\tag{2}$$ We will need both of these equations later.
If we differentiate $$y' = -\frac{x}{25y}$$ implicitly with respect to $x$, we obtain \begin{align*} y'' & = -\frac{25y - 25xy'}{625y^2} && \text{by the Quotient Rule}\\ & = -\frac{y - xy'}{25y^2} && \text{cancel a factor of $25$}\\ & = -\frac{y - x\left(-\dfrac{x}{25y}\right)}{25y^2} && \text{use equation (2) to substitute $-\dfrac{x}{25y}$ for $y'$}\\ & = -\frac{\dfrac{25y^2 + x^2}{25y}}{25y^2} && \text{add fractions in numerator}\\ & = -\frac{25y^2 + x^2}{625y^3} && \text{multiply numerator and denominator by $25y$}\\ & = -\frac{100}{625y^3} && \text{use equation (1) to substitute $100$ for $x^2 + 25y^2$}\\ & = -\frac{4}{25y^3} && \text{cancel a factor of $25$} \end{align*}