What is the derivative of #f(x)=(4x-1)/x#?

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Answer 1 · 2016-09-15T20:54:43

#(1) / (x^(2))#

We have: #f(x) = (4 x - 1) / (x)#

This function can be differentiated using the "quotient rule":

#=> f'(x) = ((x) (4) - (4 x - 1) (1)) / ((x)^(2))#

#=> f'(x) = (4 x - 4 x + 1) / (x^(2))#

#=> f'(x) = (1) / (x^(2))#

Answer 2 · 2016-09-15T23:16:31

#f'(x)=1/x^2#

Alternatively, we can split up the fraction:

#f(x)=(4x-1)/x=(4x)/x-1/x=4-x^-1#

Use the power rule, #d/dxx^n=nx^(n-1)#, to differentiate this. Recall that the derivative of the constant is #0#.

#f'(x)=0-(-x^-2)=1/x^2#