How to evaluate $\frac{d}{dx} (3x + 1)^{\frac{3}{2}}(2x + 4)$

Question

I'm supposed to use the product rule to differentiate: $$ \frac{d}{dx} (3x + 1)^{\frac{3}{2}}(2x + 4). $$

This gives me: $$ 2(3x+1)^{\frac{3}{2}} + \frac{9}{2}(2x + 4)(3x + 1)^{\frac{1}{2}}. $$

My problem is that i'm now supposed to somehow factorise this answer to match the one in the back of the book and I don't know how: $$ 5(3x+1)^{\frac{1}{2}} (3x + 4). $$

The fractional indicies are confusing me the most when it comes to factorizing, please help me. ^_^

Thank you for your time~!

Answer 1

$$2\color{red}{(3x+1)^{\frac{3}{2}}} + \frac{9}{2}(2x + 4)(3x + 1)^{\frac{1}{2}}=2\color{red}{(3x+1)}\color{blue}{(3x+1)^{\frac{1}{2}}}+ \frac{9}{2}(2x + 4)\color{blue}{(3x + 1)^{\frac{1}{2}}}$$ Factor $(3x + 1)^{\frac{1}{2}}$ out, $$=\color{blue}{(3x + 1)^{\frac{1}{2}}}[(6x+2)+(9x+18)]=(3x+1)^{\frac{1}{2}}5(3x + 4).$$

Answer 2

First notice that: $$\frac{d}{dx}(ax+b)^n=an(ax+b)^{n-1}$$ Let $p=(3x+1)^{\frac32}\to p'=3(\frac32)(3x+1)^{\frac32-1}=\frac92(3x+1)^\frac12$

Let $q=2x+4\to q'=2$

Then via product rule: $$\frac{d(pq)}{dx}=p'q+q'p$$

The letters are interchangeable, I like $p$ and $q$, others like $u$ and $v$, etc

We get: $$\frac{d[(3x+1)^\frac32(2x+4)]}{dx}=\frac92(3x+1)^\frac12(2x+4)+2(3x+1)^\frac32$$ We can then simplify this to: $$(3x+1)^\frac12\bigg[9(x+2)+2(3x+1)\bigg]$$ $$\to(3x+1)^\frac12(15x+20)\to5(3x+1)^\frac12(3x+4)$$