# Math Help - Differentiation- Quotient Rule

1. ## Differentiation- Quotient Rule

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

2. If your question is to calculate the derivative of the function here is step by step

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

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

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

can now conclude

3. Since you will need to use the chain rule on the square root anyway, you might find it easier to do this as $f(x)= (x+2)(3x+ 1)^{-1/2}$ and use the product rule rather than the quotient rule.

4. Thanks to both of you with your reply, I will try it out!!

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

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

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

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

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

I will appreciate if you are able to show me the workings to simplfying $\frac{dy}{dx}$. Thanks.

6. Hello! Just a little update on my workings...

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

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

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

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

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

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

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

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