Question

If x^3 + y^3 = 9, find the value of d^2y/dx^2 at the point (2,1)
The value of d^2y/dx^2 at the point (2,1) is

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Answer 1

${\bf{Given:-\::}}$

The equation given as,

$\bf{x^3\:+\:y^3\:=\:9}$

$\\ {\bf{To\: Find\::}}$

$\bf{\dfrac{d^2y}{dx^2}\:at\:point\:(2\:,\:1)}$

$\\ {\bf{Solution:-\::}}$

$:\implies\:\tt{x^3\:+\:y^3\:=\:9}$

$:\implies\:\tt{y^3\:=\:9\:-\:x^3}$

$:\implies\:\tt{y\:=\:\left(9\:-\:x^3\right)^{\frac{1}{3}}}$

Now,

Differentiate the above equation w.r.t x, we get

$:\implies\:\tt{\dfrac{dy}{dx}\:=\:\dfrac{d\left(9\:-\:x^3\right)^{\frac{1}{3}}}{dx}\:}$

$:\implies\:\tt{\dfrac{dy}{dx}\:=\:\dfrac{d\left(9\:-\:x^3\right)^{\frac{1}{3}}}{d\left(9\:-\:x^3\right)}\times{\dfrac{d\left(9\:-\:x^3\right)}{dx}}\:}$

$:\implies\:\tt{\dfrac{dy}{dx}\:=\:\dfrac{1}{3}\:\left(9\:-\:x^3\right)^{\frac{1}{3}\:-\:1}\times{\left(0\:-\:3\:x^{3\:-\:1}\right)}\:}$

$:\implies\:\tt{\dfrac{dy}{dx}\:=\:\dfrac{1}{3}\:\left(9\:-\:x^3\right)^{-\frac{2}{3}}\times{-3\:x^{2}}\:}$

$:\implies\:\tt{\dfrac{dy}{dx}\:=\:-\:x^2\:\left(9\:-\:x^3\right)^{-\frac{2}{3}}\:}$

Again,

Differentiate the above equation w.r.t x, we get

$:\implies\:\tt{\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\:=\:\dfrac{d}{dx}\bigg(-\:x^2\:\left(9\:-\:x^3\right)^{-\frac{2}{3}}\bigg)\:}$

$:\implies\:\tt{\dfrac{d^2y}{dx^2}\:=\:-\:x^2\:\bigg[\dfrac{d}{dx}\left(\left(9\:-\:x^3\right)^{-\frac{2}{3}}\right)\bigg]\:+\:\bigg[\dfrac{d}{dx}\:\left(-\:x^2\right)\bigg]\:\left(9\:-\:x^3\right)^{-\frac{2}{3}}}$

$:\implies\:\tt{\dfrac{d^2y}{dx^2}\:=\:-\:x^2\times{-\dfrac{2}{3}}\left(9\:-\:x^3\right)^{-\frac{2}{3}\:-\:1}\times{\left(0\:-\:3x^{3\:-\:1}\right)}\:+\:\bigg\{-\:2x^{2\:-\:1}\left(9\:-\:x^3\right)^{-\frac{2}{3}}\bigg\}\:}$

$:\implies\:\tt{\dfrac{d^2y}{dx^2}\:=\:-\:x^2\times{-3x^2}\times{-\dfrac{2}{3}}\left(9\:-\:x^3\right)^{-\frac{5}{3}}\:+\:\bigg\{-\:2x\:\left(9\:-\:x^3\right)^{-\frac{2}{3}}\bigg\}\:}$

$:\implies\:\tt{\dfrac{d^2y}{dx^2}\:=\:-\:2x^4\times\:\left(9\:-\:x^3\right)^{-\frac{5}{3}}\:-\:2x\:\left(9\:-\:x^3\right)^{-\frac{2}{3}}\:}$

Putting the value of 'x = 2' in the above equation, we get

$:\implies\:\tt{\left(\dfrac{d^2y}{dx^2}\right)_{[at\:(2\:,\:1)]}\:=\:-\:2\times{2}^4\times\:\left(9\:-\:2^3\right)^{-\frac{5}{3}}\:-\:2\times{2}\:\left(9\:-\:2^3\right)^{-\frac{2}{3}}\:}$

$:\implies\:\tt{\left(\dfrac{d^2y}{dx^2}\right)_{[at\:(2\:,\:1)]}\:=\:-\:2\times{16}\times\:\left(9\:-\:8\right)^{-\frac{5}{3}}\:-\:4\:\left(9\:-\:8\right)^{-\frac{2}{3}}\:}$

$:\implies\:\tt{\left(\dfrac{d^2y}{dx^2}\right)_{[at\:(2\:,\:1)]}\:=\:-\:32\times\:\times{1}^{-\frac{5}{3}}\:-\:4\times{1}^{-\frac{2}{3}}\:}$

$:\implies\:\tt{\left(\dfrac{d^2y}{dx^2}\right)_{[at\:(2\:,\:1)]}\:=\:-\:32\:-\:4\:}$

$:\implies\:\bf\pink{\left(\dfrac{d^2y}{dx^2}\right)_{[at\:(2\:,\:1)]}\:=\:-\:36\:}$