generalized version of Therorem 4

$\displaystyle F(x) = \int_1^{\ln ({x^2})} {\frac{1} {t}dt}$
generalized theorem: $\displaystyle F(x) = \int_{h(x)}^{g(x)} {f(t)dt} $$\displaystyle F'(x) = f(g(x))*g'(x) = f(h(x))*h'(x) I tired: \displaystyle \frac{1} {{\ln ({x^2})}}*\frac{{2x}} {{{x^2}}} = \frac{{2x}} {{{x^2}\ln ({x^2})}} \displaystyle F'(e) = \frac{{2e}} {{{e^2}\ln {{(e)}^2}}} = \frac{{2e}} {{2{e^2}}} = .368 is this correct? • Oct 4th 2009, 10:39 AM redsoxfan325 Quote: Originally Posted by genlovesmusic09 \displaystyle F(x) = \int_1^{\ln ({x^2})} {\frac{1} {t}dt} generalized theorem: \displaystyle F(x) = \int_{h(x)}^{g(x)} {f(t)dt}$$\displaystyle F'(x) = f(g(x))*g'(x) = f(h(x))*h'(x)$
$\displaystyle \frac{1} {{\ln ({x^2})}}*\frac{{2x}} {{{x^2}}} = \frac{{2x}} {{{x^2}\ln ({x^2})}}$
$\displaystyle F'(e) = \frac{{2e}} {{{e^2}\ln {{(e)}^2}}} = \frac{{2e}} {{2{e^2}}} = .368$
Yes, but your professor will probably want it written as $\displaystyle \frac{1}{e}$ as opposed to a rounded decimal.