`y = (ln(x))/(x^2)` Find y' and y''.

Textbook Question

Chapter 3, 3.6 - Problem 24 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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kalau | (Level 2) Adjunct Educator

Posted on

 Find y' and y''.

An easier way to solve for their derivatives is by rewriting this so that we do not need to use the quotient rule.

`y= ln(x)/x^2 = (x^-2) ln(x)`

We can use product rule to solve for y'.  The product rule is: AB'+BA'

`y' =x^-2 * (1/x)+ ln(x) * (-2x^-3)`

With regards to negative exponents, we can rewrite them as a fraction.

`y' =1/x^2 * (1/x)+ ln(x) * (-2/x^3)= 1/x^3-2ln(x) /x^3`

The answer to the first derivative is:  `y'= (1-2ln(x))/x^3`

For the second derivative, we will take the derivative of the first derivative.  Rewrite the first derivative so that we can use product rule instead of the quotient rule for simplicity.

`y'=(x^-3) (1-2ln(x))`

Use the product rule AB'+BA' to find the second derivative.

`y''=(x^-3)(-2/x)+(1-2ln(x))(-3x^-4)`

Rewrite the negative exponents to avoid confusion.

`y''=(1/x^3)(-2/x)+(1-2ln(x))(-3/x^4)`

Simplify.

`y''=(1/x^3)(-2/x)-3(1-2ln(x))(1/x^4)`

Since the denominators are like, we can deal with both numerators and combine the terms.  Be careful with signs!  We are adding a quantity when we distribute the negative three through the 1-2ln(x)!!!

`y''=-2/x^4 +(-3+6ln(x))/x^4`

Combine denominators.

`y''= (-2-3+6ln(x))/x^4` 

The answer to the second derivative is:  `y''= (6ln(x)-5) /x^4`

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to find the first derivative of the function, using the quotient rule, such that:

`y' = ((ln x)'*x^2 - ln x*(x^2)')/((x^2)^2)`

`y' = (x^2/x - 2x*lnx)/(x^4)`

`y' = (x - 2x*lnx)/(x^4)`

You need to evaluate the second derivative, differentiating the first derivative, with respect to x, such that:

`y'' = ((x - 2x*lnx)'(x^4) - (x - 2x*lnx)(x^4)')/((x^4)^2)`

`y'' = ((1 - 2*lnx - (2x)/x)(x^4) - 4x^3(x - 2x*lnx))/(x^8)`

`y'' = ((1 - 2*lnx - 2)(x^4) - 4x^3(x - 2x*lnx))/(x^8)`

Factoring out x^3, yields:

`y'' = x^3*(x*(1 - 2*lnx - 2) - 4(x - 2x*lnx))/(x^8)`

Reducing like terms, yields:

`y'' = ((1 - 2*lnx - 2) - 4(x - 2x*lnx))/(x^5) = (6lnx - 5)/(x^5)`

Hence, evaluating the first and the second derivatives, yields `y' = (x - 2x*lnx)/(x^4)` and `y'' = (6lnx - 5)/(x^5)`

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