Derivative of Integral With Limits

By the Fundamental Theorem of Calculus, you have $$f'(x)=\sqrt{1+x^3}.$$

Alternatively recall that $x=f(f^{-1}(x))$, so differentiating both sides of the equation and using the chain rule we have $$1=f'(f^{-1}(x))(f^{-1}(x))'$$

so $$(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}=\frac{1}{\sqrt{1+(f^{-1}(x))^3}}$$

then since $f(3)=0$ we have $f^{-1}(0)=3$ (as you mentioned) and thus

$$(f^{-1}(0))'=\frac{1}{\sqrt{1+(f^{-1}(0))^3}}=\frac{1}{\sqrt{1+27}}=\frac{1}{\sqrt{28}}.$$

Tags:

Integration

Calculus

Derivatives

Inverse Function

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