<s>[INST] 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?
 [/INST] 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 $x = 5$
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