# Math Help - generalized version of Therorem 4

1. ## generalized version of Therorem 4

$F(x) = \int_1^{\ln ({x^2})} {\frac{1}
{t}dt}
$

generalized theorem: $F(x) = \int_{h(x)}^{g(x)} {f(t)dt}
$
$F'(x) = f(g(x))*g'(x) = f(h(x))*h'(x)
$

I tired:
$\frac{1}
{{\ln ({x^2})}}*\frac{{2x}}
{{{x^2}}} = \frac{{2x}}
{{{x^2}\ln ({x^2})}}
$

$F'(e) = \frac{{2e}}
{{{e^2}\ln {{(e)}^2}}} = \frac{{2e}}
{{2{e^2}}} = .368
$

is this correct?

2. Originally Posted by genlovesmusic09
$F(x) = \int_1^{\ln ({x^2})} {\frac{1}
{t}dt}
$

generalized theorem: $F(x) = \int_{h(x)}^{g(x)} {f(t)dt}
$
$F'(x) = f(g(x))*g'(x) = f(h(x))*h'(x)
$

I tired:
$\frac{1}
{{\ln ({x^2})}}*\frac{{2x}}
{{{x^2}}} = \frac{{2x}}
{{{x^2}\ln ({x^2})}}
$

$F'(e) = \frac{{2e}}
{{{e^2}\ln {{(e)}^2}}} = \frac{{2e}}
{{2{e^2}}} = .368
$

is this correct?
Yes, but your professor will probably want it written as $\frac{1}{e}$ as opposed to a rounded decimal.