Using FTC, find $F'(x)$ where $F$ is defined on $[0,\infty)$ by $F(x)=\int_x^{e^x}\sqrt{1+t^2}dt.$

Question

Using FTC, find $F'(x)$ where $F$ is defined on $[0,\infty)$ by $$F(x)=\int_x^{e^x}\sqrt{1+t^2}dt.$$

I know $$F(x)=\int_x^{e^x}\sqrt{1+t^2}dt= \int_0^{e^x}\sqrt{1+t^2}dt-\int_0^{^x}\sqrt{1+t^2}dt$$

Then $F'(x)=\sqrt{1+(e^x)^2}-=\sqrt{1+(x)^2}$

Right?

Answer 1

No. By the chain rule the function $\int_0^{e^x}\sqrt{1+t^2}dt$ has the derivative $\sqrt{1+(e^x)^2}e^x$.Hence

$F'(x)=\sqrt{1+(e^x)^2}e^x-\sqrt{1+x^2}$.

Answer 2

With $u=\ln(t)$ you have $\text{d}u=\text{d}t/t=e^{-u}\text{d}t$ Then $$ G\left(x\right)=\int_{0}^{e^x}\sqrt{1+t^2}\text{d}t=\int_{-\infty}^{x}\sqrt{1+e^{2u}}e^{u}\text{d}u $$

And then $$ G'\left(x\right)=\sqrt{1+e^{2x}}e^{x} $$ The other part is good

Answer 3

HINT: the first derivative is given by $$-1/2\,x\sqrt {{x}^{2}+1}-1/2\,{\rm arcsinh} \left(x\right)+1/2\,{ {\rm e}^{x}}\sqrt {{{\rm e}^{2\,x}}+1}+1/2\,{\rm arcsinh} \left({ {\rm e}^{x}}\right) $$