# Thread: derivative of t of an integral wrt x

1. ## derivative of t of an integral wrt x

this is the derivative WRT t of the integral from 1 to t of f(x) WRT x

WRT=with respect to

this is asking 'what is the derivative or rate of change WRT t of the function you took the derivative of WRT x to get 1/(x^3+x+1)'

am i reading this right? and if so what is the procedure to solve it?

I think anti-derive the expression (x^3+x+1)^(-1) WRT x and then evaluate that at t and then take it's derivative WRT t.

Am I on the right track?

also how would I find the anti-derivative of $\displaystyle (x^3+x+1)^-1$ since the derivative of 1/x is ln|x| would it be $\displaystyle ln|3x^2+1|$?

thanks!

2. ## Re: derivative of t of an integral wrt x

Integrate and evaluate at the limits. Since the upper limit is t, the result of the integration will be a function of t. Then differentiate.

3. ## Re: derivative of t of an integral wrt x

Originally Posted by kingsolomonsgrave
Yours as well as reply #2 is a pure waste of time.

If each of $\displaystyle f~\&~g$ is a differentiable function then:
$\displaystyle \frac{d}{{dt}}\left[ {\int_{g(t)}^{f(t)} {h(x)dx} } \right] = f'(t)h(f(t)) - g'(t)h(g(t))$

4. ## Re: derivative of t of an integral wrt x

Plato is right. What you need to know is the Fundamental Theorem of Calculus. Plato gave you the general formula, but this one is the one you'll use most of the time:

$\displaystyle \frac{d}{dt}\int_c^th(x)dx=h(t)$

where c is a constant.

- Hollywood