# What is the value of f^'(2)?

## Given, $f \left(x\right) = {\int}_{1}^{{x}^{3}}$(1/(1+lnt))dt

Apr 3, 2018

$f ' \left(2\right) = \frac{12}{1 + \ln 8}$

#### Explanation:

We have:

$f \left(x\right) = {\int}_{1}^{{x}^{3}} \frac{1}{1 + \ln t} \mathrm{dt}$

By the chain rule

$f ' \left(x\right) = \frac{3 {x}^{2}}{1 + \ln \left({x}^{3}\right)}$

Thus

$f ' \left(2\right) = \frac{3 {\left(2\right)}^{2}}{1 + \ln \left(8\right)}$

$f ' \left(2\right) = \frac{12}{1 + \ln 8}$

Hopefully this helps!

Apr 3, 2018

$\frac{12}{1 + \ln 8} = \frac{12}{\ln} \left(8 e\right)$.

#### Explanation:

Suppose that, $\int \frac{1}{1 + \ln t} \mathrm{dt} = g \left(t\right) + C$.

:. g'(t)=1/(1+lnt)...[" the Definition of Integral]"...(ast).

Further, by the Fundamental Theorem of Calculus,

$f \left(x\right) = {\int}_{1}^{{x}^{3}} \frac{1}{1 + \ln t} \mathrm{dt} = {\left[g \left(t\right) + C\right]}_{1}^{{x}^{3}} = g \left({x}^{3}\right) - g \left(1\right)$.

$\Rightarrow f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left\{g \left({x}^{3}\right) - g \left(1\right)\right\}$,

=g'(x^3)*d/dx(x^3)-0......[because," The Chain Rule]".

$= 3 {x}^{2} \cdot g ' \left({x}^{3}\right)$,

$= 3 {x}^{2} \cdot \frac{1}{1 + \ln {x}^{3}} \ldots \ldots \ldots \ldots \left[\because , \left(\ast\right)\right]$,

$\therefore f ' \left(x\right) = \frac{3 {x}^{2}}{1 + \ln {x}^{3}}$.

$\Rightarrow f ' \left(2\right) = \frac{3 \cdot {2}^{2}}{1 + \ln {2}^{3}} = \frac{12}{\ln e + \ln 8} = \frac{12}{\ln} \left(8 e\right)$.