# Differentiation- Quotient Rule

• Feb 10th 2010, 04:15 AM
Punch
Differentiation- Quotient Rule
Calculate the gradient of the tangent to the curve $\displaystyle y=\frac{x+2}{\sqrt{3x+1}}$ at $\displaystyle x=1$
• Feb 10th 2010, 04:31 AM
danielomalmsteen
If your question is to calculate the derivative of the function here is step by step

$\displaystyle y=\frac{x+2}{\sqrt{3x+1}} \rightarrow y'=\frac{(x+2)'(\sqrt{3x+1}) - (\sqrt{3x+1})'(x+2)}{3x+1}$

The derivate of $\displaystyle \sqrt{3x+1}$ is calculated using the chain rule

$\displaystyle f(x)=\sqrt{3x+1} = (3x+1)^{\frac{1}{2}} \rightarrow f'(x) = \frac{1}{2}\cdot \frac{1}{\sqrt{3x+1}} \cdot 3$

can now conclude
• Feb 10th 2010, 04:50 AM
HallsofIvy
Since you will need to use the chain rule on the square root anyway, you might find it easier to do this as $\displaystyle f(x)= (x+2)(3x+ 1)^{-1/2}$ and use the product rule rather than the quotient rule.
• Feb 11th 2010, 05:21 AM
Punch
Thanks to both of you with your reply, I will try it out!!
• Feb 11th 2010, 07:05 AM
Punch
Quote:

Originally Posted by danielomalmsteen
If your question is to calculate the derivative of the function here is step by step

$\displaystyle y=\frac{x+2}{\sqrt{3x+1}} \rightarrow y'=\frac{(x+2)'(\sqrt{3x+1}) - (\sqrt{3x+1})'(x+2)}{3x+1}$

The derivate of $\displaystyle \sqrt{3x+1}$ is calculated using the chain rule

$\displaystyle f(x)=\sqrt{3x+1} = (3x+1)^{\frac{1}{2}} \rightarrow f'(x) = \frac{1}{2}\cdot \frac{1}{\sqrt{3x+1}} \cdot 3$

can now conclude

Hi, I am well aware of the formula to use, the problem lies in simplifying the gradient function $\displaystyle \frac{dy}{dx}$...

I will appreciate if you are able to show me the workings to simplfying $\displaystyle \frac{dy}{dx}$. Thanks.
• Feb 11th 2010, 06:38 PM
Punch
Hello! Just a little update on my workings...

Quote:

Calculate the gradient of the tangent to the curve $\displaystyle y=\frac{x+2}{\sqrt{3x+1}}$ at $\displaystyle x=1$
$\displaystyle Let U=x+2$ and $\displaystyle V=\sqrt{3x+1}$

$\displaystyle \frac{dU}{dx}=1$

$\displaystyle \frac{dV}{dx}=\frac{1}{2}(3x+1)^{-\frac{1}{2}}(3)$

$\displaystyle =\frac{3}{2}(3x+1)^{-\frac{1}{2}}$

$\displaystyle \frac{dy}{dx}=\frac{\frac{3x+1-(\frac{3}{2}(x+2))}{\sqrt{3x+1}}}{3x+1}$

$\displaystyle =\frac{(3x+1)^2-(\frac{3}{2}(x+2)(3x+1))}{\sqrt{3x+1}}$

$\displaystyle =\frac{3x+1(3x+1-((\frac{3}{2})(x+2)))}{\sqrt{3x+1}}$

I can't simplify further. Am I on the right track?