Finding derivative of the function by using the definition of the derivative

Find the derivative of the function by using the definition of the derivative.

f(x)=(x+1)/(x-1)

REMEMBER: The first derivative  IS  the gradient of the tangent to the curve

Basically when you find this from first principals you take the gradient of the secant.

I want the gradient of the tangent at (x,f(x))

So I have found the gradient of the SECANT  from  (x,f(x)) to the point   (x+c, f(x+c))

THEN I find the limit of this as c tends to 0

This limit will be  the gradient of the TANGENT at (x,f(x)).

In my case this is

    (That is not written technically at all)

This can be expressed in a couple of different ways.  This is one of them.

\(f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{f(x+c)-f(x)}{(x+c)-x}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{\frac{x+c+1}{x+c-1}-\frac{x+1}{x-1}}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{\frac{(x+c+1)(x-1)}{(x+c-1)(x-1)}-\frac{(x+1)(x+c-1)}{(x-1)(x+c-1)}}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{\frac{(x+c+1)(x-1)-(x+1)(x+c-1)}{(x+c-1)(x-1)}}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{\frac{x(x+c+1)-1(x+c+1) -x(x+c-1)-1(x+c-1)}{(x+c-1)(x-1)}}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{\frac{2x-2(x+c) }{(x+c-1)(x-1)}}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{-2c}{(x+c-1)(x-1)}\times \frac{1}{c}\\ f'(x)=\displaystyle \lim _{c\rightarrow 0} \frac{-2}{(x+c-1)(x-1)}\\ f'(x)=\frac{-2}{(x-1)^2} \)

Check using quotient rule.

f(x)=(x+1)/(x-1)

\(f(x)=\frac{(x+1)}{(x-1)}\\ f'(x)=\frac{(x-1)*1-(x+1)*1}{(x-1)^2}\\ f'(x)=\frac{-2}{(x-1)^2}\\\)

Great :)

I hope you take the time needed to actually learn from my answer