Let $h(x)=\int_2^{10/x}arctan(t)dt$. Use the fundamental Theorem of Calculus to find $h'(x)$.

So I believe the answer is $arctan(10/x)-arctan(2)$, but am being told it is incorrect. Any other ideas?

My thought process was that the derivative of an integral will give back the integrand, then evaluate $F(b)-F(a)$ where F is the original integrand.


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    $\begingroup$ Apply the Fundamental Theorem of Calculus (and apply chain rule) $\endgroup$ – Morgan Rodgers Jan 24 at 0:43

if $F(x) - F(b) = \int_b^{x} f(t) \ dt$

$F'(x) = f(x)$

This is the fundamental theorem of calculus.

We have

$F(\frac {10}{x}) - F(b) = \int_b^{\frac{10}{x}} f(t) \ dt$

$\frac {d}{dx}\left(F(\frac {10}{x}) - F(b)\right) = \frac {d}{dx}\int_b^{\frac{10}{x}} f(t) \ dt$

Remember, b is a constant, so $\frac {d}{dx} F(b) = 0$

For the other term we are going to use the chain rule.

$F'(\frac {10}{x}) = f(\frac {10}{x})(\frac {d}{dx} \frac {10}{x}) = f(\frac{10}{x})(\frac {-10}{x^2})$

$-\frac {10\arctan \frac {10}{x}}{x^2}$


It's true that if $F(x)=\int_a^x f(t)\,dt$, then $F'(x)=f(x)$. But in your case you have $$ h(x)=F(\tfrac{10}x), $$ where $F(x)=\int_2^x\arctan(t)\,dt$. So to find $h'$ you need to apply the Chain Rule and the Fundamental Theorem of Calculus.


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