How do you find the second derivative by implicit differentiation?

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How do you find the second derivative by implicit differentiation?

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Answer 1 · 2014-09-12T23:23:27

Let us find ${d^2y}/{dx^2}$ for $x^3+y^3=1$ .

First, let us find ${dy}/{dx}$ .
$x^3+y^3=1$
by differentiating with respect to $x$ ,
$Rightarrow 3x^2+3y^2{dy}/{dx}=0$
by subtracting $3x^2$ ,
$Rightarrow3y^2{dy}/{dx}=-3x^2$
by dividing by $3y^2$ ,
$Rightarrow {dy}/{dx}=-{x^2}/{y^2}$

Now, let us find ${d^2y}/{dx^2}$ .
by differentiating with respect to $x$ ,
$Rightarrow{d^2y}/{dx^2}=-{2x cdot y^2-x^2 cdot 2y{dy}/{dx}}/{(y^2)^2} =-{2x(y^2-xy{dy}/{dx})}/{y^4}$
by plugging in ${dy}/{dx}=-{x^2}/{y^2}$ ,
$Rightarrow{d^2y}/{dx^2}=-{2x[y^2-xy(-x^2/y^2)]}/y^4=-{2x(y^2+x^3/y)}/y^4$
by multiplying the numerator and the denominator by $y$ ,
$Rightarrow{d^2y}/{dx^2}=-{2x(y^3+x^3)}/y^5$
by plugging in $y^3+x^3=1$ ,
$Rightarrow{d^2y}/{dx^2}=-{2x}/y^5$

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How do you find the second derivative by implicit differentiation?

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