Differentiating $x^2 \sqrt{2x+5}-6$ How do I differentiate this function: f(x)= $x^2  \sqrt{2x+5}-6$
I had: I had $2x\sqrt{2x+5} + x^2  \dfrac{1}{2\sqrt{2x+5}}$ but the correction model said it was I had $2x\sqrt{2x+5} + x^2  \dfrac{2}{2\sqrt{2x+5}}$
 A: $$
\begin{eqnarray*}
  y &=& x^2(2x+5)^{1/2} - 6 \\
  \frac{dy}{dx} &=& \frac{d}{dx} \Big[ x^2(2x+5)^{1/2} \Big] - \frac{d}{dx}\Big[ 6 \Big], \qquad \textrm{Sum/Difference Rule}\\
  &=& \frac{d}{dx}\Big[x^2\Big](2x+5)^{1/2} + x^2\frac{d}{dx}\Big[(2x+5)^{1/2}\Big] - 0, \qquad \textrm{Product Rule}\\
  &=& 2x(2x+5)^{1/2} + x^2\left( \frac{1}{2}(2x+5)^{-1/2}\cdot 2\right),
  \qquad \textrm{Chain Rule}\\
  &=& 2x\sqrt{2x+5} + \frac{ x^2}{\sqrt{2x+5}}, \qquad \textrm{simplification.}
\end{eqnarray*}
$$
A: Work from the outside in. Begin by differentiating it term by term: $$f\,'(x)=\frac{d}{dx}\left(x^2\sqrt{2x+5}\right)-\frac{d}{dx}(6)=\frac{d}{dx}\left(x^2\sqrt{2x+5}\right)-0\;.$$
Now you have to calculate the derivative of $x^2\sqrt{2x+5}$. This is a product, so you use the product rule:
$$\left[x^2\sqrt{2x+5}\right]'=x^2\left[\sqrt{2x+5}\right]'+\left[x^2\right]'\sqrt{2x+5}\;.$$
To complete the differentiation you’ll need the derivative of $x^2$, which is very easy, and the derivative of $\sqrt{2x+5}$. That one is also pretty easy once you rewrite the function as $(2x+5)^{1/2}$: the power rule and the chain rule will take care of it.
A: Using the chain rule where $\cfrac {df}{dx} = \cfrac {dg}{du} \cfrac {du}{dx} $ if $f(x) = g(u(x))$ and the product rule where $\cfrac {d}{dx} (uv) = v\cfrac {du}{dx} + u\cfrac {dv}{dx} $
A: $2x . \sqrt{2x+5} + \dfrac{x^2}{2} . \dfrac{1}{\sqrt{2x+5}}2$
A: I would use the product rule, which you seem to have tried from your comment above. Here would be the idea:
$\begin{split}
\frac{d[x^2 \sqrt{2x+5} - 6]}{dx}
 &= \frac{dx^2}{dx} \sqrt{2x+5} +  x^2 \frac{d(2x+5)^{\1/2}}{dx} \\
 &= 2x \sqrt{2x+5} + x^2 \frac{1}{2} (2x+5)^{-1/2} \cdot 2 \\
 &= 2x \sqrt{2x+5} + \frac{x^2}{\sqrt{2x+5}}
\end{split}
$
