# Thread: prduct & Quotient Rule - Differentiation

1. ## prduct & Quotient Rule - Differentiation

Please could somebody look at my attachehment and let me know if im ok with what I have done.

Thanks

Recall chain rule: $\displaystyle h(x)=f(g(x))\implies h'(x)=f'(g(x))\cdot g'(x).$

Recall product rule: $\displaystyle (f(x)\cdot g(x))'=f'(x)\cdot g(x)+f(x)\cdot g'(x).$

Recall quotient rule: $\displaystyle \left( {\frac{{f(x)}} {{g(x)}}} \right)' = \frac{{f'(x) \cdot g(x) - f(x) \cdot g'(x)}} {{g^2 (x)}}.$

Try it again.

3. Sorry, Richard!

They are all wrong . . .

You don't have a grasp on the Product Rule and Quotient Rule, do you?

. . $\displaystyle \begin{array}{cccccccc}\text{Producr Rule:}& y & = & f(x)\cdot g(x) &\Longrightarrow& y' &=& f(x)\cdot g'(x) + f\,'(x)\cdot g(x) \\ \\ \text{Quotient Rule:} & y &=& \dfrac{f(x)}{g(x)}& \Longrightarrow & y' &=& \dfrac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{[g(x)]^2} \end{array}$

$\displaystyle (a)\;\;y \;=\;\overbrace{x}^{{\color{blue}f(x)}}\cdot\overb race{e^{2x}}^{{\color{blue}g(x)}}\qquad\text{(Prod uct)}$
$\displaystyle y' \;=\;\overbrace{x}^{{\color{blue}f(x)}}\cdot\overb race{2e^{2x}}^{{\color{blue}g'(x)}} + \overbrace{1}^{{\color{blue}f'(x)}}\cdot\overbrace {e^{2x}}^{{\color{blue}g(x)}} \;=\;2xe^{2x} + e^{2x} \;=\;\boxed{e^{2x}(2x+1)}$

$\displaystyle (b)\;\;y \;=\;\overbrace{t^{\frac{1}{2}}}^{{\color{blue}f(x )}}\cdot \overbrace{\cos t\,}^{{\color{blue}g(x)}}\qquad\text{(Product)}$
$\displaystyle y' \;=\;\overbrace{\left(t^{\frac{1}{2}}\right)}^{{\c olor{blue}f(x)}} \overbrace{(-\sin t)}^{{\color{blue}g'(x)}} + \overbrace{\left(\frac{1}{2}t^{-\frac{1}{2}}\right)}^{{\color{blue}f'(x)}} \overbrace{ (\cos t)}^{{\color{blue}g(x)}} \;=\;\boxed{-t^{\frac{1}{2}}\sin t + \frac{\cos t}{2t^{\frac{1}{2}}}}$

$\displaystyle (c)\;\;y \;=\;\frac{\overbrace{\sin x}^{{\color{blue}f(x)}}}{\underbrace{x}_{{\color{b lue}g(x)}}}\qquad\text{(Quotient)}$
$\displaystyle y' \;=\;\frac{\overbrace{x}^{{\color{blue}g(x)}} \cdot \overbrace{\cos x}^{{\color{blue}f'(x)}} - \overbrace{\sin x}^{{\color{blue}f(x)}} \cdot \overbrace{1}^{{\color{blue}g'(x)}} } {\underbrace{x^2}_{{\color{blue}[g(x)]^2}} } \;=\;\boxed{\frac{x\cos x - \sin x}{x^2}}$

4. Thanks