Inverse integration

Calculus Level 4

Let f ( x ) = ∫ 2 x t 1 + t 4 dt f(x) = \displaystyle \int_2^x \dfrac{t}{\sqrt{1 + t^4}}\text{dt} and g be the inverse of f f .
Then the value of g ′ ( 0 ) g'(0) is

Clarification :

  • f ′ ( x ) f'(x) means first derivative of f ( x ) f(x) with respect to x.
None of these 0 0 17 17 17 \sqrt{17}

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Akhil Bansal
Oct 31, 2015

If g is the inverse of f, then, ⇒ g o f ( x ) = x ⇒ g ( f ( x ) ) = x \Rightarrow g o f (x) = x \Rightarrow g(f(x))= x Differentiating both sides with respect to x.
g ′ ( f ( x ) ⋅ f ′ ( x ) = 1 g'(f(x) \cdot f'(x) = 1 On observing, f ( 2 ) = 0 f(2) = 0 .
g ′ ( f ( 2 ) ) ⋅ f ′ ( 2 ) = 0 ⇒ g ′ ( 0 ) ⋅ f ′ ( 2 ) = 1 g'(f(2)) \cdot \color{#3D99F6}{f'(2)} = 0 \Rightarrow \color{#D61F06}{g'(0)} \cdot\color{#3D99F6}{ f'(2) }= 1
Now finding f ′ ( x ) f'(x) from given integral using second fundamental theorem of calculus,
f ′ ( x ) = x 1 + x 4 f'(x) = \dfrac{x}{\sqrt{1 + x^4}} f ′ ( 2 ) = 2 17 \color{#3D99F6}{f'(2)} = \dfrac{2}{\sqrt{17}}

g ′ ( 0 ) = 17 2 \color{#D61F06}{g'(0)} = \dfrac{\sqrt{17}}{2}

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