Question

Can I get help answering this question? It's a derivatives question, further info in the attachment.

Answer 1

$\bf f(x)=|3+9x|\implies f(x)=\sqrt{(3+9x)^2}\implies f(x)=[(3+9x)^2]^{\frac{1}{2}} \\\\\\ \cfrac{dy}{dx}=\stackrel{chain~rule}{\cfrac{1}{2}[(3+9x)^2]^{-\frac{1}{2}}\cdot 2(3+9x)\cdot 9}\implies \cfrac{dy}{dx}=\cfrac{9(3+9x)}{[(3+9x)^2]^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\cfrac{27+81x}{|3+9x|}\\\\ -------------------------------$

$\bf \textit{left-hand derivative at }x=-\frac{3}{9}\implies \cfrac{27+81x}{-(3+9x)} \\\\\\ \cfrac{27+81\left( -\frac{3}{9} \right)}{-3-9\left( -\frac{3}{9} \right)}\implies \cfrac{27-27}{-3+3}\implies \stackrel{unde fined}{\cfrac{0}{0}}\\\\ -------------------------------\\\\ \textit{right-hand derivative at }x=-\frac{3}{9}\implies \cfrac{27+81x}{+(3+9x)} \\\\\\ \cfrac{27+81\left( -\frac{3}{9} \right)}{3+9\left( -\frac{3}{9} \right)}\implies \cfrac{27-27}{3-3}\implies \stackrel{unde fined}{\cfrac{0}{0}}$