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

Answer 1

GIVEN :–

$\\ \bf \implies x^3 + y^3 = 9$

TO FIND :–

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

SOLUTION :–

$\\ \bf \implies x^3 + y^3 = 9$

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

$\\ \bf \implies y = (9 - {x}^{3})^ \frac{1}{3}$

• Differentiate with respect to 'x' –

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

$\\ \bf \implies \dfrac{dy}{dx} = - {x}^{2} (9 - {x}^{3})^ {- \frac{2}{3}}$

• Again Differentiate with respect to 'x' –

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

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

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

• At (2,1) –

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

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

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

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

$\\ \implies \red{ \boxed{ \bf\left(\dfrac{d^{2} y}{dx ^{2}}\right)_{(2,1)}=-36}}$

Answer 2

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

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

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

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

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

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

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