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How do you differentiate #y = ((2x+3)^4)/ x#?

Answer 1

Aurora Alvarado

#(dy)/(dx)=(3(2x-1)(2x+3)^3)/x^2#

Here,

#y=(2x+3)^4/x#

Quotient Rule is used to obtain

#(dy)/(dx)=(xd/(dx)((2x+3)^4)-(2x+3)^4d/(dx)(x))/(x)^2#

#=>(dy)/(dx)=(x*4(2x+3)^3 xx2-(2x+3)^4*1)/x^2#

#=>(dy)/(dx)=((2x+3)^3{8x-2x-3})/x^2#

#=>(dy)/(dx)=((2x+3)^3(6x-3))/x^2#

#=>(dy)/(dx)=(3(2x+3)^3(2x-1))/x^2#

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Answer 2

Evelyn Agard

To differentiate the function ( y = \frac{{(2x+3)^4}}{x} ), we use the quotient rule, which states that if ( u ) and ( v ) are differentiable functions of ( x ), then the derivative of ( \frac{u}{v} ) is given by ( \frac{u'v - uv'}{v^2} ).

Let ( u = (2x+3)^4 ) and ( v = x ). Then, ( u' ) (the derivative of ( u )) is ( 4(2x+3)^3 \cdot 2 ) and ( v' ) (the derivative of ( v )) is ( 1 ).

Substituting into the quotient rule formula, we get:

[ y' = \frac{(4(2x+3)^3 \cdot 2 \cdot x) - ((2x+3)^4 \cdot 1)}{x^2} ]

[ y' = \frac{8x(2x+3)^3 - (2x+3)^4}{x^2} ]

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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.