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I've got a pretty simple derivative question for you guys. Currently, I'm a high school shop teacher preparing kids for a timed calculus competition. It's been almost 45 years since I've taken multivariate calculus, and I'm having a hard time remembering some tricks.

Almost every round features where you must find $\frac{dy}{dx}$ given something like this.

$$ x^2 + 2xy +7 = 0 $$

The kids are taught to differentiate implicitly, then solve for $\frac{dy}{dx}$. I remembered that you could set the original equation equal to some function $g$, and simplify with this formula (from chain rule): $$\frac{dy}{dx} = \frac{-\frac{\partial g}{\partial x}}{\frac{\partial g}{\partial y}}$$

I know there's a way to extend this to finding the second derivative of $y$ with respect to $x$, but I just cannot remember how.

I think I take the partial of the first derivative wrt $x$, multiply by the first derivative, then add the partial of the first derivative wrt $y$, but it doesn't seem to be working.

Any help?

Przemysław Scherwentke
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Mike Jones
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2 Answers2

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If $$ y'=-\frac{F_x}{F_y}, $$ it can be computed, that $$ y''=-\frac{F'_xF_y-F_xF'_y}{F^2_y}=-\frac{(F_{xx}+F_{xy}y')F_y-F_x(F_{yx}+F_{yy}y')}{F^2_y}=\\\\=-\frac{\left( F_{xx}+F_{xy}\left( -\frac{F_x}{F_y}\right)\right) F_y-F_x\left( F_{yx}+F_{yy}\left( -\frac{F_x}{F_y}\right) \right) }{F^2_y}=\\\\=-\frac{F_y^2 F_{xx}-F_xF_y(F_{xy}+F_{yx})+F_x^2F_{yy}}{F^3_y} $$

However, $y=y(x)$, hence differentiating $y'(x)$ is usually a quicker method than using the formula above.

Przemysław Scherwentke
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It's just the chain rule, so it doesn't need to be as complicated as the solution above.

If you call your first derivative $h$, you can find the second derivative with this.

$$ \frac{d^2y}{dx^2}=\frac{\partial h}{\partial x} + \frac{\partial h}{\partial y}\frac{dy}{dx} $$

What you said at the end was almost correct, but you had backward which partial you multiplied by the first derivative.

user162226
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