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Compute the derivative of a function defined by an integral equation

Compute the derivative f'(x) for the function f(x) defined as

    \[ f(x) = \int_{x^3}^{x^2} \frac{t^6}{1+t^4} \, dt \]

without evaluating the integral directly.


Let

    \[ A(x) = \int_0^x \frac{t^6}{1+t^4} \, dt. \]

Then, by the fundamental theorem of calculus,

    \[ A'(x) = \frac{x^6}{1+x^4}. \]

So, using the linearity of the integral with respect to the interval of integration we write,

    \begin{align*}    f(x) &= \int_{x^3}^{x^2} \frac{t^6}{1+t^4} \, dt \\  &= \int_0^{x^2} \frac{t^6}{1+t^4} \, dt - \int_0^{x^3} \frac{t^6}{1+t^4} \, dt \\  &= A(x^2) - A(x^3). \end{align*}

Now, differentiating and using the chain rule we have

    \begin{align*}  f'(x) &= (A(x^2))' - (A(x^3))' \\  &= (2x) A'(x^2) - (3x^2) A'(x^3) \\  &= (2x) \frac{x^{12}}{1+x^8} - (3x^2) \frac{x^{18}}{1+x^{12}} \\  &= \frac{2x^{13}}{1+x^8} - \frac{3x^{20}}{1+x^{12}}. \end{align*}

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