#2 What is dy/dx if y = log(x+1/x).

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mitch1 | (Level 1) Adjunct Educator

Posted on

Solution:

You can answer the problem by rewriting the equation first. That is by using the properties of logarithm. 

y = log(x + 1/x)

y = log((x^2+1)/x)

y = log(x^2+1)  - log x

Now, you have to recall the derivative formula  for logarithm.That is,

log y = 1/y * dy

dy/dx = 1/(x^2+1) *derivative of (x^2+1)  - 1/x* derivative of (x)

dy/dx = 1/(x^2+1) * 2x  - 1/x * 1

dy/dx= 2x/(x^2+1) -  1/x 

Simplify your answer by performing the subtraction operation.

LCD = x(x^2+1)

dy/dx = 2x*(x)/(x(x^2+1)) - 1*(x^2+1)/(x(x^2+1))

dy/dx = 2x^2/(x(x^2+1))  -x^2/(x(x^2+1)) - 1/(x(x^2+1))

dy/dx = (x^2-1)/(x(x^2+1))

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The derivative of the function `y = log(x + 1/x)` with respect to x, `dy/dx` has to be determined. It is assumed that log in the problem refers to natural logarithm.

Use the chain rule here. Let `z = x + 1/x` and y = log z.

`dz/dx = 1 - 1/x^2` and `dy/dz = 1/z`

Multiplying the two derivatives:

`(dz/dx)*(dy/dz) = (1 - 1/x^2)*(1/z)`

=> `dy/dx = (1 - 1/x^2)*(1/z)`

Substituting `z = x + 1/x`

=> `dy/dx = (1 - 1/x^2)/(x + 1/x)`

The derivative of the function `y = log(x + 1/x)` ` `with respect to x is `dy/dx = (1 - 1/x^2)/(x + 1/x)`

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