Question

Please help!
How do I find the derivative of:
the integral from 1 to sqrt(x) inverse tan(x) dx

Answer 1

\[d/dx\int\limits_{1}^\sqrt{x}\tan^{-1}(x)dx\]
That it?

Answer 2

Yep!

Answer 3

it's inverse tan of x

Answer 4

part of the fundamental theorem of calculus, the derivative of an an integral is the function that's being integrated

Answer 5

From the fundamental theorem of calculus we know that
\[g(x)=\int\limits_{a=x}^{t=x}f(t)dt\]
where a is a fixed a number

and the upper limit of the integral varies

\[g'(x)=f(x)\]
Or the original functon f(t) evaluated at x

For example
\[g(x)=\int\limits_{1}^{x}\sqrt{1-t^2}dt\]
\[g'(x)=\sqrt{1-x^2}\]

When you take the derivative of an integral and the upper limit isn't just x (for example 2x, 3x, x^2) you must use the chain rule. We use the chain rule because we use substitution.
\[g(x)=\int\limits_{1}^{x^2}\sin(t)dt\]
We substitute x^2 = u
\[g'(x)=d/dx \int\limits_{1}^{u}\sin(t)dt\]
\[g'(x)=(d/du \int\limits_{1}^{u}\sin(t)dt) du/dx\]
According to the fundamental theorem of calculus
\[d/du\int\limits_{1}^{u}\sin{t}dt=\sin(u)\]
This multiplied by du/dx
gives
\[du/dx(\sin(u))=2xsin(x^2)\]
Now back to the original problem, using these methods
\[d/dx\int\limits_{1}^{x^2}\tan^{-1}(t)dt\]
\[2xtan^{-1}(x^2)\]

Answer 6

-.- I put the upper limit as x^2

Answer 7

Thank you this is very helpful! However, why is the upper bound equal to x^(2)

Answer 8

Mistake on my part

Answer 9

okay. thanks again!