Calculus problem #2148

Calculus Level 2

If f ( x ) = ( x + 1 ) 2 ( 2 x + 1 ) 3 f(x)=\frac{(x+1)^2}{(2x+1)^3} , what is the value of f ′ ( 1 ) f'(1) ?

Details and assumptions

f ′ ( x ) f'(x) denotes the derivative of f ( x ) f(x) .


The answer is -0.148148.

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Arron Kau by Arron Kau

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1 solution

Leonardo Chandra
Dec 25, 2013

Use the derivative theorem for division to get: d d x \frac{d}{dx} u ( x ) v ( x ) \frac{u(x)}{v(x)} = ( u ′ ( x ) . v ( x ) − u ( x ) . v ′ ( x ) ) ( v ( x ) ) 2 \frac{(u'(x).v(x) - u(x).v'(x))}{(v(x))^2} So: u ′ ( x ) = 2 ( x + 1 ) u'(x)= 2(x+1) and v ′ ( x ) = 3.2 ( 2 x + 1 ) 2 v'(x)= 3.2(2x+1)^2

Then: f ′ ( x ) = 2 ( x + 1 ) ∗ ( 2 x + 1 ) 3 − ( x + 1 ) 2 ∗ 6 ( 2 x + 1 ) 2 ( 2 x + 1 ) 6 f'(x)= \frac{2(x+1)*(2x+1)^3 -(x+1)^2*6(2x+1)^2}{(2x+1)^6}

f ′ ( 1 ) = − 4 27 f'(1)= \frac{-4}{27}

f ′ ( 1 ) = − 0.148 f'(1)= -0.148

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