Differentiation with square roots - check my work =)

Question

Question: Differentiate √ x + 3x  (all is a square root)

I did function of a function rule

= (x²+3x)^0.5 '

= 0.5(x² + 3x) (2x+3)
= (2x + 3) ÷ 2(x² + 3x)

Is that correct?

Thanks

Answer 1

Your work is nearly correct. What Archis is referring to is your use of the power rule. Remember, the power rule for derivatives says $\frac{d}{du}u^{n} = nu^{n-1}$. So when you take the derivative of $(x^{2}+3x)^{1/2}$ you will get

\begin{align} \frac{d}{dx}(x^{2}+3x)^{1/2} &= \frac{1}{2}\cdot (x^{2}+3x)^{\color{red}{1/2-1}}\cdot(2x+3) \\ &= \frac{1}{2}\cdot (x^{2}+3x)^{\color{red}{-1/2}}\cdot(2x+3) \\ &= \frac{2x+3}{2(x^{2}+3x)^{\color{red}{1/2}}} \end{align}

I've indicated in $\color{red}{\text{red}}$ the part that is different from your answer.