• Nov 15th 2009, 05:28 PM
masterofcheese
the problem is x arctan squareroot(x)

i solve it out and got (square root of x)/2(1+x)

However, the book answer gives it as my answer + arctan(square root of x)

How did the book get the + arctan(squre root of x)

I went through my work 4 times and I can't find out why. Unless the x in fron of arctan square root of x plays a role via chain rule.

If the chain rule is the case, what happens when there is a constant, such as 29 instead of x, so the problem is 29 arctan (squareroot of x).

Would it thus have a + 29 arctan (square root of x)?
• Nov 15th 2009, 05:40 PM
skeeter
Quote:

Originally Posted by masterofcheese
the problem is x arctan squareroot(x)

i solve it out and got (square root of x)/2(1+x)

However, the book answer gives it as my answer + arctan(square root of x)

How did the book get the + arctan(squre root of x)

I went through my work 4 times and I can't find out why. Unless the x in fron of arctan square root of x plays a role via chain rule.

$\textcolor{red}{y = x\arctan{\sqrt{x}}}$

product rule ...

$\textcolor{red}{y' = x \cdot \frac{1}{2\sqrt{x}(1 + x)} + \arctan{\sqrt{x}} \cdot 1}
$

...
• Nov 15th 2009, 05:40 PM
Jose27
Remember it's a product and $(fg)'=f'g+fg'$ where $f(x)=x$ and $g(x)=\arctan (\sqrt{x} )$