Integral (d/dx)∫from(x^(1/2)) to x^2 of tan(9t) dt

[vysero]

Homework Statement

(d/dx)∫from(x^(1/2)) to x^2 of tan(9t) dt

Homework Equations

The Attempt at a Solution

2x((tan(9x^2))-(tan(x^(1/2)))

I am not sure what I am doing wrong here.

 

[Dick]

Homework Statement

(d/dx)∫from(x^(1/2)) to x^2 of tan(9t) dt

Homework Equations

The Attempt at a Solution

2x((tan(9x^2))-(tan(x^(1/2)))

I am not sure what I am doing wrong here.

You are using the fundamental theorem of calculus here, right? Can you explain how you got that? There's a few parts of right things in there. But overall, it's a mess.

 

[micromass]

Homework Statement

(d/dx)∫from(x^(1/2)) to x^2 of tan(9t) dt

Homework Equations

The Attempt at a Solution

2x((tan(9x^2))-(tan(x^(1/2)))

I am not sure what I am doing wrong here.

Your post is very hard to read. Please consider typing up your equations with LaTeX. Here is a short guide: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3 It would really help us a lot.

 

[Mark44]

Also, there is no such word as "intergral" in English. I am editing your title.

 

[HallsofIvy]

I would write the original function as
$$\int_{x^{1/2}}^{x^2} tan(9t)dt= \int_{x^{1/2}}^a tan(9t)dt+ \int_a^{x^2} tan(9t) dt$$
$$= \int_a^{x^2} tan(9t)dt- \int_a^{x^{1/2}} tan(9t)dt$$

NOW use the Fundamental Theorem of Calculus. It looks to me like you forgot the derivative of the "$x^{1/2}$" term.