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)?

2. 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.

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

product rule ...

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

3. Remember it's a product and $\displaystyle (fg)'=f'g+fg'$ where $\displaystyle f(x)=x$ and $\displaystyle g(x)=\arctan (\sqrt{x} )$