How do you differentiate #g(x)=(3x-1)/(2x+1)#?

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Answer 1 · 2018-05-06T04:03:51

I tried this:

We could use the Quotient Rule; given a function as:

#f(x)=g(x)/(h(x))#

the derivative will be:

#f'(x)=[g'(x)h(x)-g(x)h'(x)]/[h(x)]^2#

In your case:

#g'(x)=[3(2x+1)-2(3x-1)]/(2x+1)^2#

#g'(x)=[cancel(6x)+3cancel(-6x)+2]/(2x+1)^2=5/(2x+1)^2#