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

Beginner2022-07-24Added 19 answers

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

$\Rightarrow 3{x}^{2}+3{y}^{2}\frac{dy}{dx}=0$

by subtracting $3{x}^{2}$

$\Rightarrow 3{y}^{2}\frac{dy}{dx}=-3{x}^{2}$

by dividing by $3{y}^{2}$

$\Rightarrow \frac{dy}{dx}=-\frac{{x}^{2}}{{y}^{2}}$

Now, let us find $\frac{{d}^{2}y}{{dx}^{2}}$

by differentiating with respect to x

$\Rightarrow \frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x\cdot {y}^{2}-{x}^{2}\cdot 2y\frac{dy}{dx}}{{\left({y}^{2}\right)}^{2}}=-\frac{2x({y}^{2}-xy\frac{dy}{dx})}{{y}^{4}}$

by plugging in $\frac{dy}{dx}=-\frac{{x}^{2}}{{y}^{2}}$

$\Rightarrow \frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x[{y}^{2}-xy(-\frac{{x}^{2}}{{y}^{2}})]}{{y}^{4}}=-\frac{2x({y}^{2}+\frac{{x}^{3}}{y})}{{y}^{4}}$

by multiplying the numerator and the denominator by y

$\Rightarrow \frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x({y}^{3}+{x}^{3})}{{y}^{5}}$

by plugging in ${y}^{3}+{x}^{3}=1$

$\Rightarrow \frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x}{{y}^{5}}$

$\frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x}{{y}^{5}}$

First, let us find $\frac{dy}{dx}$

${x}^{3}+{y}^{3}=1$

by differentiating with respect to x

$\Rightarrow 3{x}^{2}+3{y}^{2}\frac{dy}{dx}=0$

by subtracting $3{x}^{2}$

$\Rightarrow 3{y}^{2}\frac{dy}{dx}=-3{x}^{2}$

by dividing by $3{y}^{2}$

$\Rightarrow \frac{dy}{dx}=-\frac{{x}^{2}}{{y}^{2}}$

Now, let us find $\frac{{d}^{2}y}{{dx}^{2}}$

by differentiating with respect to x

$\Rightarrow \frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x\cdot {y}^{2}-{x}^{2}\cdot 2y\frac{dy}{dx}}{{\left({y}^{2}\right)}^{2}}=-\frac{2x({y}^{2}-xy\frac{dy}{dx})}{{y}^{4}}$

by plugging in $\frac{dy}{dx}=-\frac{{x}^{2}}{{y}^{2}}$

$\Rightarrow \frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x[{y}^{2}-xy(-\frac{{x}^{2}}{{y}^{2}})]}{{y}^{4}}=-\frac{2x({y}^{2}+\frac{{x}^{3}}{y})}{{y}^{4}}$

by multiplying the numerator and the denominator by y

$\Rightarrow \frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x({y}^{3}+{x}^{3})}{{y}^{5}}$

by plugging in ${y}^{3}+{x}^{3}=1$

$\Rightarrow \frac{{d}^{2}y}{{dx}^{2}}=-\frac{2x}{{y}^{5}}$

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