Question

39-42 Find \(y''\) by implicit differentiation/ Simplify where possible.

42. \({x^{\bf{3}}} - {y^{\bf{3}}} = {\bf{7}}\)

Short answer

The value of \(y''\) is \( - \frac{{14x}}{{{y^5}}}\).

Step-by-step solution

Differentiate the given equation with respect to x

Differentiate the equation \({x^3} - {y^3} = 7\) on both sides.

\(\begin{aligned}{x^3} - {y^3} &= 7\\3{x^2} - 3{y^2}\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}x}}} \right) &= 0\\\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= \frac{{{x^2}}}{{{y^2}}}\end{aligned}\)

Find the value of \(y''\)

Find the value of \(y''\) or \(\frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}}\) by differentiating the equation \(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = \frac{{{x^2}}}{{{y^2}}}\).

\(\begin{aligned}\frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{{x^2}}}{{{y^2}}}} \right)\\ &= \frac{{{y^2}\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{x^2}} \right) - {x^2}\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{y^2}} \right)}}{{{y^4}}}\\ &= \frac{{{y^2}\left( {2x} \right) - {x^2}\left( {2y} \right)\frac{{{\rm{d}}y}}{{{\rm{d}}x}}}}{{{y^4}}}\end{aligned}\)

Solve further,

\(\begin{aligned}\frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} &= \frac{{2x{y^2} - {x^2}\left( {2y} \right)\left( {\frac{{{x^2}}}{{{y^2}}}} \right)}}{{{y^3}}}\\ &= \frac{{2x\left( {{y^3} - {x^3}} \right)}}{{{y^5}}}\\ &= \frac{{2x\left( { - 7} \right)}}{{{y^5}}}\\ &= - \frac{{14x}}{{{y^5}}}\end{aligned}\)

So, the value of \(y''\) is \( - \frac{{14x}}{{{y^5}}}\).