Find second derivative of `4x^3 + 3y^3 = 7`

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lfryerda | High School Teacher | (Level 2) Educator

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Another way of finding the derivative of `4x^3+3y^3=7` is to use implicit differentiation.  This gives:

`12x^2+9y^2{dy}/{dx}=0`   now solve for `{dy}/{dx}`

`9y^2{dy}/{dx}=-12x^2`   divide by `9y^2`   Call this equation (1)

`{dy}/{dx}=-{12x^2}/{9y^2}`

To get the second derivative, we differentiate (1) and use the product rule on the left side

`18y({dy}/{dx})^2+9y^2{d^2y}/{dx^2}=-24x`   move terms to right side

`9y^2{d^2y}/{dx^2}=-24x-18y({dy}/{dx})^2`   sub for first derivative

`9y^2{d^2y}/{dx^2}=-24x-18y(-{12x^2}/{9y^2})^2`   simplify

` ` `{d^2y}/{dx^2}=-{24x}/{9y^2}-{18y}/{9y^2}(-{12x^2}/{9y^2})^2`

`{d^2y}/{dx^2}=-{8x}/{3y^2}-{2}/{9y}cdot {16x^4}/{9y^4}`

` ` `{d^2y}/{dx^2}=-{8x}/{3y^2}-{32x^4}/{81y^5}`

`{d^2y}/{dx^2}=-{8x(27y^3+4x^3)}/{81y^5}`

The second derivative is `{d^2y}/{dx^2}=-{8x(27y^3+4x^3)}/{81y^5}` .

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to isolate `y^3 ` to the left side such that:

`3y^3 = 7 - 4x^3 => y^3 = (7 - 4x^3)/3 => y = root(3)((7 - 4x^3)/3) => y = ((7 - 4x^3)/3)^(1/3)`

Differentiating wth respect to x yields:

`dy/dx = (1/3)((7 - 4x^3)/3)^(1/3-1)((7 - 4x^3)/3)'`

`dy/dx = (1/3)((7 - 4x^3)/3)^(-2/3)(-4x^2)`

You need to differentiate again using the product rule such that:

`(d^2y)/(dx^2) = (1/3)(-2/3)((7 - 4x^3)/3)^(-2/3-1)(-4x^2)(-4x^2) + (1/3)((7 - 4x^3)/3)^(-2/3)(-8x)`

`(d^2y)/(dx^2) = (-2/9)((7 - 4x^3)/3)^(-5/3)(16x^2)+ (1/3)((7 - 4x^3)/3)^(-2/3)(-8x)`

Hence, evaluating the second derivative yields:

`(d^2y)/(dx^2) = (-2/9)((7 - 4x^3)/3)^(-5/3)(16x^2)+ (1/3)((7 - 4x^3)/3)^(-2/3)(-8x)`

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