If `ln x = y^2 + 3x` what is the value of `(d^2y)/dx^2`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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Actually, moving the terms from one side of equal to the other side changes the sign of the term that has been moved.

`y^2 = lnx - 3x`

Taking square root both sides yields: `y=+-sqrt(ln x - 3x)`

First time differentiating yields: `(dy)/(dx) = (ln x - 3x)/(2sqrt(ln x - 3x))`

Second time differentiating yields: `(d^2y)/(dx^2) = (ln x - 3x)'*2sqrt(ln x - 3x) - (ln x - 3x)*(2sqrt(ln x - 3x))/(4(ln x - 3x))`

`(d^2y)/(dx^2) = ((1/x - 3)*2sqrt(ln x - 3x) - (ln x - 3x)*((1/x - 3)/(4sqrt(ln x - 3x))))/(4(ln x - 3x))`

`(d^2y)/(dx^2) = ((1/x - 3)*(2sqrt(ln x - 3x) - (ln x - 3x)/4sqrt(ln x - 3x)))/(4(ln x - 3x))`

`` `(d^2y)/(dx^2) = (1/x - 3)*(8(ln x - 3x) - (ln x - 3x))/(16(ln x - 3x)sqrt(ln x - 3x))`

`` `(d^2y)/(dx^2) = (7(ln x - 3x)(1/x - 3))/(16(ln x - 3x)sqrt(ln x - 3x))`

`` `(d^2y)/(dx^2) = (7(1/x - 3))/(16sqrt(ln x - 3x))`

Second time differentiating yields: `(d^2y)/(dx^2) = (7(1/x - 3))/(16sqrt(ln x - 3x))`

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