# How do you use the second fundamental theorem of Calculus to find the derivative of given int dt/sqrt( 1 + t^2)  from [x, 2x]?

Jul 3, 2016

$\frac{2}{\sqrt{1 + 4 {x}^{2}}} - \frac{1}{\sqrt{1 + {x}^{2}}}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({\int}_{x}^{2 x} \frac{\mathrm{dt}}{\sqrt{1 + {t}^{2}}}\right)$

the FTC part 2 tells us that

if $F \left(x\right) = {\int}_{a}^{x} \quad f \left(t\right) \quad \mathrm{dt}$

then $\frac{\mathrm{dF}}{\mathrm{dx}} = f \left(x\right)$

so we first need to arrange the integral so that FTC-2 can be applied

${\int}_{x}^{2 x} \frac{\mathrm{dt}}{\sqrt{1 + {t}^{2}}}$

$= \textcolor{red}{{\int}_{a}^{2 x} \frac{\mathrm{dt}}{\sqrt{1 + {t}^{2}}}} - \setminus \textcolor{g r e e n}{{\int}_{a}^{x} \frac{\mathrm{dt}}{\sqrt{1 + {t}^{2}}}}$

the green term is now good to go as it matches exactly the patterm given above. we can say that

$\frac{d}{\mathrm{dx}} \left({\int}_{a}^{x} \frac{\mathrm{dt}}{\sqrt{1 + {t}^{2}}}\right) = \frac{1}{\sqrt{1 + {x}^{2}}} q \quad \star$

the term in red needs a little more work

if we re-write it as

${\int}_{a}^{\textcolor{red}{u}} \frac{\mathrm{dt}}{\sqrt{1 + {t}^{2}}}$ where $u = 2 x$ then we can again conclude from FTC-2 that

$\frac{d}{\mathrm{du}} {\int}_{a}^{\textcolor{red}{u}} \frac{\mathrm{dt}}{\sqrt{1 + {t}^{2}}} = f \left(u\right)$

then applying the chain rule

$\frac{d}{\mathrm{dx}} {\int}_{a}^{\textcolor{red}{u}} \frac{\mathrm{dt}}{\sqrt{1 + {t}^{2}}}$

$= \frac{d}{\mathrm{du}} {\int}_{a}^{\textcolor{red}{u}} \frac{\mathrm{dt}}{\sqrt{1 + {t}^{2}}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= f \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= \frac{1}{\sqrt{1 + {\left(2 x\right)}^{2}}} \cdot 2$

combining this with $\star$ gives

$\frac{2}{\sqrt{1 + 4 {x}^{2}}} - \frac{1}{\sqrt{1 + {x}^{2}}}$