Why is my answer different from the book's in differentiating this function?

[-Dragoon-]

Homework Statement

Differentiate the following function: f(x) = x\sqrt{x^{3}+2x}

Homework Equations

N/A

The Attempt at a Solution

Here's my attempt:
First I rewrite it as:
f(x) = x(x^{3}+2x)^{\frac{1}{2}}
Now, I differentiate using chain rule:
f(x) = x(3x{^2}+2x)(\frac{1}{2})(x^{3}+2x)^{\frac{1}{2}-1}
By simplifying and rearranging the equation, I get:
f(x) = \frac{3x^{3}+2x}{2\sqrt{x^{3}+2x}}
But, this what my book got:
f(x) = \frac{3x^{3}+2x}{2\sqrt{x^{3}+2x}} + \sqrt{x^{3}+2x}

I checked my answer over thrice just in case I made any minor errors, and I don't think I have. Where did I go wrong? Did I apply the wrong rule? Thanks in advance.

 

[Disconnected]

This problem calls for both the chain rule, as there is a function within a function, and the product rule, as the overall function is the product of g(x)=x and h(x)=(x^3+2x)^(1/2)

 

[-Dragoon-]

This problem calls for both the chain rule, as there is a function within a function, and the product rule, as the overall function is the product of g(x)=x and h(x)=(x^3+2x)^(1/2)

I see what I did wrong. Thanks for the help.

 

[mahdisadjadi]

Well,
<br /> f(x)=x(x^{3}+2x)^{\frac{1}{2}}<br />
If we apply the product rule, we should write:
<br /> \frac{df(x)}{dx}=(\frac{dx}{dx})[(x^{3}+2x)^{\frac{1}{2}}]+x(\frac{d[(x^{3}+2x)^{\frac{1}{2}}]}{dx})<br />
So if we rearrange the equations:
<br /> \frac{df(x)}{dx}=(x^{3}+2x)^{\frac{1}{2}}+\frac{1}{2}\frac{3x^{3}+2x}{\sqrt{x^{3}+2x}}<br />

But it we tend to apply the chain rule, we should write for example:
<br /> f(x)=xg(x) \Rightarrow df(x)=x dg(x)+ g(x) dx<br />

Since g is a gunction of x, we should apply both the chain rule and the product rule.so,
<br /> \frac{df(x)}{dx}=x\frac{dg(x)}{dx}+g(x)\frac{dx}{dx}=x\frac{dg(x)}{dx}+g(x)=\frac{1}{2}\frac{3x^{3}+2x}{\sqrt{x^{3}+2x}}+(x^{3}+2x)^{\frac{1}{2}}<br />

 

[-Dragoon-]

Thank you for that thorough explanation, mahdisadjadi.