# How do you find the second derivative of y^2=x^3?

Jul 29, 2015

y''= (3)/(4sqrtx)

#### Explanation:

You can use implicit differentiation:

$D \left({y}^{2}\right) = D \left({x}^{3}\right)$

$2 y y ' = 3 {x}^{2}$

$y ' = \frac{3 {x}^{2}}{2 y}$

Use the quotient rule:

$y ' ' = \frac{2 y .6 x - 3 {x}^{2} .2 y '}{4 {y}^{2}}$

Subs for $y ' \Rightarrow$

$y ' ' = \frac{12 x y - 6 {x}^{2} \left(3 {x}^{2} / \left(2 y\right)\right)}{4 {y}^{2}}$

$y ' ' = \frac{12 x y - 9 {x}^{4} / y}{4 {y}^{2}}$

$y ' ' = \frac{12 x {y}^{2}}{4 {y}^{3}} - \frac{9 {x}^{4}}{4 {y}^{3}}$

${y}^{2} = {x}^{3} \Rightarrow$

$y ' ' = \frac{12 {x}^{4}}{4 {y}^{3}} - \frac{9 {x}^{4}}{4 {y}^{3}}$

$y ' ' = \frac{3 {x}^{4}}{4 {y}^{3}}$

Since ${y}^{2} = {x}^{3}$ this becomes:

$y ' ' = \frac{3 {x}^{4}}{4 y {x}^{3}}$

$y ' ' = \frac{3 x}{4 y}$

${y}^{2} = {x}^{3}$

$y = {x}^{\frac{3}{2}}$

Substituting for y gives:

$y ' ' = \frac{3 x}{4 {x}^{\frac{3}{2}}}$

$y ' ' = \frac{3}{4 \sqrt{x}}$

You can also use explicit differentiation:

${y}^{2} = {x}^{3}$

$y = \sqrt{{x}^{3}}$

$y = {x}^{\frac{3}{2}}$

Apply the power rule:

$y ' = \frac{3}{2.} {x}^{\frac{1}{2}}$

And again for the 2nd derivative:

$y ' ' = \frac{3}{4} {x}^{- \frac{1}{2}}$

y''=3/4.(1)/(x^(1/2)

$y ' ' = \frac{3}{4 \sqrt{x}}$

Either way gets you there.