Calculate the gradient of the tangent to the curve $\displaystyle y=\frac{x+2}{\sqrt{3x+1}}$ at $\displaystyle x=1$
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
Hello! Just a little update on my workings...
$\displaystyle Let U=x+2$ and $\displaystyle V=\sqrt{3x+1}$Calculate the gradient of the tangent to the curve $\displaystyle y=\frac{x+2}{\sqrt{3x+1}}$ at $\displaystyle x=1$
$\displaystyle \frac{dU}{dx}=1$
$\displaystyle \frac{dV}{dx}=\frac{1}{2}(3x+1)^{-\frac{1}{2}}(3)$
$\displaystyle =\frac{3}{2}(3x+1)^{-\frac{1}{2}}$
$\displaystyle \frac{dy}{dx}=\frac{\frac{3x+1-(\frac{3}{2}(x+2))}{\sqrt{3x+1}}}{3x+1}$
$\displaystyle =\frac{(3x+1)^2-(\frac{3}{2}(x+2)(3x+1))}{\sqrt{3x+1}}$
$\displaystyle =\frac{3x+1(3x+1-((\frac{3}{2})(x+2)))}{\sqrt{3x+1}}$
I can't simplify further. Am I on the right track?