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