# Checking on an answer – Finding the derivative of a square root.

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• January 12th 2010, 06:45 PM
ckoeber
Checking on an answer – Finding the derivative of a square root.
So, I am working on finding the derivative of the following function:
http://www.churchleadership.com/temp/Problem2.png
Using the chain rule I did the following:
http://www.churchleadership.com/temp/Solution2.png
My two questions are:
1. Is this right?

2. If correct, should I leave it in this form?
Thank you so much for your time.
• January 12th 2010, 07:59 PM
Dark Sun
The answer I got was:

Let $f(t)=\sqrt{t+\sqrt t}$, and let F(t) be an antiderivative of f(t). Then;

$\int _0^{\tan x}\sqrt{t+\sqrt t} dt=F(t)]_0^{\tan x}=F(\tan x)-F(0).$

So, by the chain rule:

$\frac{d}{dx}\int _0^{\tan x}\sqrt{t+\sqrt t} dt=F'(\tan x)\cdot \sec ^2 x-F'(0)=\sqrt{\tan x+\sqrt{\tan x}}\cdot \sec ^2x$.
• January 13th 2010, 08:26 AM
ckoeber
I see, you did the substitutions that I didn't do. I'll rework the problem and check back!

Thanks again.