Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. h'(x) = I got -arctan x dx/x^2..think it's wrong
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Originally Posted by Jgirl689 Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. h'(x) = I got -arctan x dx/x^2..think it's wrong If $\displaystyle h(x)=\int_a^{g(x)} f(t) dt$, then: $\displaystyle h'(x)=f\left(g(x)\right) g'(x)$. In your answer, It should be $\displaystyle arctan \left( \frac{1}{x} \right)$ not just $\displaystyle arctan(x)$.
Originally Posted by Jgirl689 Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. h'(x) = I got -arctan x dx/x^2..think it's wrong if $\displaystyle h(x) = \int_a^u f(t) \, dt$ , then $\displaystyle h'(x) = f(u) \cdot \frac{du}{dx}$ $\displaystyle h'(x) = \arctan\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)$
Let $\displaystyle \arctan t = F'(t)$ Then $\displaystyle h(x) = \int _6 ^{1/x} \arctan t dt = F(1/x) - F(6)$ $\displaystyle h'(x) = F'\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)$
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