Finding the equation of a tangent line to a curve at a given point. Find the equation for the tangent (in standard form) to the curve defined by $$y = \sqrt\frac{5x}{2x-1}$$
For the procedure of this question would I simply differentiate it using the quotient rule? From which you insert $x=5$ in to, and then use $y_2-y_1 = m(x_2-x_1)$ to find the equation.
I got this as my answer but it looks very ugly...
$y = 0.74x + 4.633$
(yes i know it isn't in standard form)
Am I doing something wrong?
 A: The point on the curve will be $(5,\frac{5}{3})$. Then differentiate $y$ using the chain rule and quotient rule
\begin{align}
\frac{dy}{dx} = \frac{1}{2}\bigg(\frac{5x}{2x - 1}\bigg)^{-\frac{1}{2}}\bigg( \frac{5(2x - 1) - (2)(5x)}{(2x - 1)^2}\bigg)
\end{align}
Then by pluggin in $x = 5$
\begin{align}
\frac{dy}{dx}\bigg|_{x=5} &= \frac{1}{2}\bigg(\frac{25}{9} \bigg)^\frac{-1}{2}\left(\frac{-5}{81}\right)\\
&= \frac{1}{2} \cdot \frac{3}{5} \cdot \frac{-5}{81} = -\frac{1}{54}
\end{align}
Use the point slope form
\begin{align}
y - \frac{5}{3} &= -\frac{1}{54}(x - 5 )\\
y + \frac{1}{54}x - \frac{95}{54} &= 0
\end{align}
Which is the equation of the tangent line at $x = 5$
A: For $$y = \sqrt\frac{5x}{2x-1}$$
The gradient $m$ can be found using the product rule $$m=\frac{d}{dx}\sqrt\frac{5x}{2x-1}=\sqrt{5}\frac{d}{dx}\left({\frac{x}{2x-1}}\right)^{\frac12}=\sqrt{5}\frac{d}{dx}\left(x^{\frac12}(2x-1)^{-\frac12}\right)=\sqrt{5}\left(\frac12x^{-\frac12}(2x-1)^{-\frac12}-x^{\frac12}(2x-1)^{-\frac32}\right)$$
At $x=5$ $$m=\sqrt{5}\left((5)^{\frac12}\cdot-(2(5)-1)^{-\frac32}+(2(5)-1)^{-\frac12}\cdot\frac12(5)^{-\frac12}\right)$$
$$=\sqrt{5}\left(-\sqrt{5}\cdot 9^{-\frac32}+9^{-\frac12}\cdot\frac{1}{2\sqrt{5}}\right)$$
$$=\sqrt{5}\left(\frac{1}{6\sqrt{5}}-\frac{\sqrt{5}}{27}\right)=-\frac{1}{54}$$
at $x=5\implies y=\sqrt{\cfrac{5(5)}{2(5)-1}}=\sqrt{\cfrac{25}{9}}=\cfrac53$
The equation of the straight line at $\big(5,\frac53 \big)$ is therefore
$$y-\frac53 = -\frac{1}{54}(x-5)\implies \color{blue}{y = \frac{95-x}{54}}$$
A: make the Ansatz $$y=-\frac{1}{54}x+n$$ and for $x=5$ we get $y=\frac{5}{3}$ thus we will have 
$$\frac{5}{3}=-\frac{5}{54}+n$$
Can you proceed?
