Find the equation of the tangent to the curve at te given point:
B) $f(x)=\frac{1}{\sqrt(x-1)}$ At $(2,1)$
I know the procedure on how to do it but I'm getting the wrong answer. First I found the slope and had to use binomial conjugate step, etc.
The answer is $x+2y-4=0$
The tangent of the curve of a function $f$ at a point $(x_0,f(x_0))$ has the equation: $$y=f'(x_0)(x-x_0)+f(x_0)$$ and in your example $f(x)=(x-1)^{-1/2}$ and $x_0=2$ hence $f'(2)=-\frac12$ so the equation is $$y=-\frac 1 2(x-2)+1\iff x+2y-4=0$$
Edited The slope of the tangent at $(2,1)$ is $$\lim_{h\to0}\frac{f(h+2)-f(2)}{h}=\lim_{h\to0}\frac{\frac{1}{\sqrt{h+1}}-1}{h}=\lim_{h\to0}\frac{(1-\sqrt{h+1})\color{red}{\times(1+\sqrt{h+1})}}{h\sqrt{h+1}\color{red}{\times(1+\sqrt{h+1})}}\\ =\lim_{h\to0}\frac{-1}{\sqrt{h+1}\color{red}{\times(1+\sqrt{h+1})}} =-\frac12$$
By differentiation we get:
$f'(x)=\frac{-(x-1)^{-3/2}}{2}$
We can then substitute $x=2$ to find the gradient of the funtion at $(2,1)$,
$f(2)=\frac{-(2-1)^{-3/2}}{2}=\frac{-1}{2}$
So the gradient is $\frac{-1}{2}$.
Now we know the slope and a point on the line, therefore we can calculate the equation of the line. The equation of a straight line is always in the form $y=mx +c$, using this fact and substituting in the gradient we just found we get:
$y=\frac{-1}{2}x+c$, luckily we also know a point on the line to substitute in to find $c$.
Using $(2,1)$:
$1=\frac{-1}{2}2+c$
rearranges to $c=2$ so, $y=\frac{-1}{2}x+2$
Multiplying everything by $2$ will give us: $2y=x+4$ or alternatively; $x+2y-4=0$
