What is the ((((2x-1)^2))/(((x+2)^3)))^' ?

The ((((2x-1)^2))/(((x+2)^3)))^' is ((2x-1)(-2x+11))/((x+2)^4)

What is the first ((((2x-1)^2))/(((x+2)^3)))^' ?

The first ((((2x-1)^2))/(((x+2)^3)))^' is ((2x-1)(-2x+11))/((x+2)^4)

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\frac{( ( 2 x - 1 ) ^{2} )}{( ( x + 2 ) ^{3} )}^{^{'}}

\frac{( 2 x - 1 ) ( - 2 x + 11 )}{( x + 2 ) ^{4}}

Solution steps

\frac{( ( 2 x - 1 ) ^{2} )}{( ( x + 2 ) ^{3} )}^{'}

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{( ( 2 x - 1 ) ^{2} ) ^{^{'}} ( x + 2 ) ^{3} - ( ( x + 2 ) ^{3} ) ^{^{'}} ( 2 x - 1 ) ^{2}}{( ( x + 2 ) ^{3} ) ^{2}}

((2 x - 1)^{2})^{'} = 4 (2 x - 1)

((x + 2)^{3})^{'} = 3 (x + 2)^{2}

= \frac{4 ( 2 x - 1 ) ( x + 2 ) ^{3} - 3 ( x + 2 ) ^{2} ( 2 x - 1 ) ^{2}}{( ( x + 2 ) ^{3} ) ^{2}}

Simplify

\frac{4 ( 2 x - 1 ) ( x + 2 ) ^{3} - 3 ( x + 2 ) ^{2} ( 2 x - 1 ) ^{2}}{( ( x + 2 ) ^{3} ) ^{2}} : \frac{( 2 x - 1 ) ( - 2 x + 11 )}{( x + 2 ) ^{4}}

= \frac{( 2 x - 1 ) ( - 2 x + 11 )}{( x + 2 ) ^{4}}

lo g_{10} (sin (x^{2}))

5 sin (x + 1)

\frac{l}{800}

\frac{5 x ^{3}}{4}

100 q + 5 q^{2}

What is the ((((2x-1)^2))/(((x+2)^3)))^' ?
The ((((2x-1)^2))/(((x+2)^3)))^' is ((2x-1)(-2x+11))/((x+2)^4)
What is the first ((((2x-1)^2))/(((x+2)^3)))^' ?
The first ((((2x-1)^2))/(((x+2)^3)))^' is ((2x-1)(-2x+11))/((x+2)^4)