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How do you find the derivative of #(x-3)/(2x+1)#?

Answer 1

Scarlett Allison

#7/(2x+1)^2#

#"differentiate using the "color(blue)"quotient rule"#

#"given "f(x)=(g(x))/(h(x))" then"#

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"#

#g(x)=x-3rArrg'(x)=1#

#h(x)=2x+1rArrh'(x)=2#

#rArrf'(x)=((2x+1)-2(x-3))/(2x+1)^2#

#color(white)(rArrf'(x))=7/(2x+1)^2#

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Answer 2

Ezra Adams

#=7/(4x^2-4x+1)# or #7/(2x+1)^2#

Quotient rule: #(u'v )-(uv')# all divided by #v^2#, assuming the top line is #u# and bottom is #v#

In this function, #u= x-3# and #v=2x+1#

#:. u' = 1 " and " v' = 2#

Thus, inserting that yields:

#(1*(2x+1)-(x-3)*2)/(2x+1)^2#

#=((2x+1)-(2x-6))/(4x^2+4x+1)#

#= (cancel(2x)+1-cancel(2x)+6)/(4x^2-4x+1)#

#=7/(4x^2-4x+1)#

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Answer 3

Aurora Alvarado

To find the derivative of ((x-3)/(2x+1)), apply the quotient rule which states that the derivative of (u/v) is ((vu' - uv')/v^2). (u = x-3) and (v = 2x+1). The derivatives are (u' = 1) and (v' = 2). Applying the quotient rule gives ((vu' - uv')/v^2 = ((2x+1)(1) - (x-3)(2))/(2x+1)^2). Simplify this expression to get the derivative.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.