# How do you use the limit definition to find the derivative of #f(x)=(x+1)/(x-1)#?

The limit definition of derivative is:

so:

By signing up, you agree to our Terms of Service and Privacy Policy

To find the derivative of ( f(x) = \frac{x+1}{x-1} ) using the limit definition, we start by using the definition of the derivative:

[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} ]

Substitute ( f(x) = \frac{x+1}{x-1} ) into the definition:

[ f'(x) = \lim_{h \to 0} \frac{\frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}}{h} ]

Next, simplify the expression inside the limit:

[ f'(x) = \lim_{h \to 0} \frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{h(x+h-1)(x-1)} ]

Expand the numerator:

[ f'(x) = \lim_{h \to 0} \frac{x^2 - x + hx - h - x + 1 - x^2 - hx - x - h + 1}{h(x+h-1)(x-1)} ]

Simplify the numerator:

[ f'(x) = \lim_{h \to 0} \frac{-2h}{h(x+h-1)(x-1)} ]

Cancel out the ( h ) in the numerator and denominator:

[ f'(x) = \lim_{h \to 0} \frac{-2}{(x+h-1)(x-1)} ]

Now, plug in ( h = 0 ):

[ f'(x) = \frac{-2}{(x-1)^2} ]

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find the equation of the tangent line to the curve #y= 4/(x-1)# at point of (0,-4)?
- What is the slope of the line normal to the tangent line of #f(x) = sec^3x+sin(2x-(3pi)/8) # at # x= (11pi)/8 #?
- How do you find an equation of the tangent line to the curve at the given point #y = cos(2x)# and #x=pi/4#?
- How do you find the equation of the line tangent to the graph of #F(x)=x^2+2x+1# at (-3,4)?
- How do you find the equation of the secant line through the points where X has the given values: #h(x)=sqrt(x-5)#; x=6, x=9?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7