Donna Flynn

2022-07-23

How do you use implicit differentiation to find $\frac{{d}^{2}y}{{dx}^{2}}$ of ${x}^{3}+{y}^{3}=1$

yermarvg

By implicitly differentiating twice, we can find
$\frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x}{{y}^{5}}$
First, let us find $\frac{dy}{dx}$
${x}^{3}+{y}^{3}=1$
by differentiating with respect to x
$⇒3{x}^{2}+3{y}^{2}\frac{dy}{dx}=0$
by subtracting $3{x}^{2}$
$⇒3{y}^{2}\frac{dy}{dx}=-3{x}^{2}$
by dividing by $3{y}^{2}$
$⇒\frac{dy}{dx}=-\frac{{x}^{2}}{{y}^{2}}$
Now, let us find $\frac{{d}^{2}y}{{dx}^{2}}$
by differentiating with respect to x
$⇒\frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x\cdot {y}^{2}-{x}^{2}\cdot 2y\frac{dy}{dx}}{{\left({y}^{2}\right)}^{2}}=-\frac{2x\left({y}^{2}-xy\frac{dy}{dx}\right)}{{y}^{4}}$
by plugging in $\frac{dy}{dx}=-\frac{{x}^{2}}{{y}^{2}}$
$⇒\frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x\left[{y}^{2}-xy\left(-\frac{{x}^{2}}{{y}^{2}}\right)\right]}{{y}^{4}}=-\frac{2x\left({y}^{2}+\frac{{x}^{3}}{y}\right)}{{y}^{4}}$
by multiplying the numerator and the denominator by y
$⇒\frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x\left({y}^{3}+{x}^{3}\right)}{{y}^{5}}$
by plugging in ${y}^{3}+{x}^{3}=1$
$⇒\frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x}{{y}^{5}}$

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