The function is $$f(x)=\int_{2}^x \sqrt{1+t^2}dt $$ Using the formula $$\frac{1}{f'(f^{-1}(a))} $$ I get $$f'(x)=\sqrt{1+x^2} $$ but do I now need to integrate the above derivative in order to find f(x), or find the inverse of the derivative?
2026-03-27 20:19:40.1774642780
Derivative of an multivariable Inverse Function
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$$\left(\int\limits_a^xg(t)dt\right)'=g(x),$$ which gives the answer immediately: $$f'(x)=\sqrt{1+x^2}.$$