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Use the fundamental theorem of calculus to find the derivative of given functions

Find the derivative f'(x) for each of the following (without attempting to evaluate the integrals)

  1. \displaystyle{f(x) = \int_0^x (1+t^2)^{-3} \, dt},
  2. \displaystyle{f(x) = \int_0^{x^2} (1+t^2)^{-3} \, dt},
  3. \displaystyle{f(x) = \int_{x^3}^{x^2} (1+t^2)^{-3} \, dt}.

For all parts of the problem let

    \[ A(x) = \int_0^x (1+t^2)^{-3} \, dt. \]

From the fundamental theorem of calculus we then know

    \[ A'(x) = (1+x^2)^{-3}. \]

  1. Now, f(x) = A(x) so

        \[ f'(x) = A'(x) = (1+x^2)^{-3}. \]

  2. In this case f(x) = A(x^2) so

        \[ f'(x) = (A(x^2))' = (2x)A'(x^2) = (2x)(1+x^4)^{-3}. \]

  3. Now, we rewrite the integral,

        \[ f(x) = \int_{x^3}^{x^2} (1+t^2)^{-3} \, dt = \int_0^{x^2} (1+t^2)^{-3} - \int_0^{x^3} (1+t^2)^{-3} \, dt. \]

    Therefore,

        \[ f(x) = A(x^2) - A(x^3) \quad \implies \quad f'(x) = (A(x^2))' - (A(x^3))'. \]

    From above we know

        \begin{align*}  (A(x^2))' = (2x) A'(x^2) &= (2x)(1+x^4)^{-3} \\  (A(x^3))' = (3x^2) A'(x^3) &= (3x^2)(1+x^6)^{-3}. \end{align*}

    Therefore,

        \[ f'(x) = (2x)(1+x^4)^{-3} - 3x^2(1+x^6)^{-3}. \]

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