How do you differentiate #f(x)=(4x)/(2x+3)^2# using the quotient rule?

Answer 1

Answer: #h'(x)=1/4(9-8x^2)/(x+2)^4 #

Quotient rule of differentiating states that, given a function #h(x)= f(x)/g(x)# then a) #h'(x)=[f'(x)g(x)-f(x)g'(x)]/[g(x)]^2# Let #f(x)=4x and g(x)= (2x+3)^2 # b) Differentiate #f(x)=>f'(x) and g(x)=>g'(x)# #f'(x) = 4; g'(x)= 4(2x+3) # c) insert b) into a) d) #h'(x)=[4*(2x+3)^2-4x*4(2x+3) ]/(2x+3)^4# Answer: #h'(x)=1/4(9-8x^2)/(x+2)^4 #

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Answer 2

Isabella Ackerman

To differentiate ( f(x) = \frac{4x}{(2x + 3)^2} ) using the quotient rule:

Identify ( u(x) = 4x ) and ( v(x) = (2x + 3)^2 ).
Apply the quotient rule: [ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} ]
Find ( u'(x) ) and ( v'(x) ) using the power rule and chain rule, respectively.
- ( u'(x) = 4 )
- ( v'(x) = 2(2x + 3)(2) = 4(2x + 3) )
Plug the values into the quotient rule formula: [ f'(x) = \frac{(4)(2x + 3)^2 - (4x)(4(2x + 3))}{(2x + 3)^4} ]
Simplify the expression to get the derivative of ( f(x) ).

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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.