How do you use the limit definition of the derivative to find the derivative of #f(x)=(x+1)/(x-4)#?

Question

Samantha Adkison · Accepted Answer

#f'(x)=(-5)/((x-4)^2)# #f'(x)=Lim_(hrarr0)(f(x+h)-f(x))/h# #f'(x)=Lim_(hrarr0)((((x+h+1)/(x+h-4))-((x+1)/(x-4)))/h)# #f'(x)=Lim_(hrarr0)((((x+h+1)(x-4)-(x+h-4)(x+1))/((x+h-4)(x-4)))/h)# #f'(x)=Lim_(hrarr0)(((x+h+1)(x-4)-(x+h-4)(x+1))/(h(x+h-4)(x-4)))# #f'(x)=Lim_(hrarr0)((x^2-4x+hx-4h+x-4)-(x^2+x+hx+h-4x-4))/(h(x^2-4x+hx-4h-4x+16))# #f'(x)=Lim_(hrarr0)(x^2-4x+hx-4h+x-4-x^2-x-hx-h+4x+4)/(h(x^2-4x+hx-4h+x-4))# #f'(x)=Lim_(hrarr0)(cancel(x^2)-cancel(4x)+cancel(hx)-4h+cancel(x)-cancel(4)-cancel(x^2)-cancel(x)-cancel(hx)-h+cancel(4x)+cancel(4))/(h(x^2-4x+hx+4h-4x+16))# #f'(x)=Lim_(hrarr0)(-5cancel(h))/(cancel(h)(x^2-4x+hx+4h-4x+16))# #f'(x)=Lim_(hrarr0)(-5)/(x^2-4x+hx+4h-4x+16)# in the limit #" "hx," "4h" "=0# #f'(x)=(-5)/(x^2-4x-4x+16)# #f'(x)=(-5)/(x^2-8x+16)# #f'(x)=(-5)/((x-4)^2)#

Evelyn Agard · Answer

undefined To use the limit definition of the derivative to find the derivative of $ f(x) = \frac{x+1}{x-4} $, you would follow these steps:

1. Write down the limit definition of the derivative:
$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

2. Substitute the given function into the formula:
$$ f'(x) = \lim_{h \to 0} \frac{\frac{(x+h)+1}{(x+h)-4} - \frac{x+1}{x-4}}{h} $$

3. Simplify the expression:
$$ f'(x) = \lim_{h \to 0} \frac{\frac{x+h+1}{x+h-4} - \frac{x+1}{x-4}}{h} $$

4. Find a common denominator and combine the fractions:
$$ f'(x) = \lim_{h \to 0} \frac{[(x+h+1)(x-4) - (x+1)(x+h-4)]}{(x+h-4)(x-4)h} $$

5. Expand and simplify the numerator:
$$ f'(x) = \lim_{h \to 0} \frac{x^2 - 3xh - 4x + xh + h - 4 - (x^2 - 3xh - 4x + x + xh - 4)}{(x+h-4)(x-4)h} $$

6. Combine like terms:
$$ f'(x) = \lim_{h \to 0} \frac{-8h}{(x+h-4)(x-4)h} $$

7. Cancel out common factors and simplify:
$$ f'(x) = \lim_{h \to 0} \frac{-8}{(x+h-4)(x-4)} $$

8. Evaluate the limit as $ h $ approaches 0:
$$ f'(x) = \frac{-8}{(x-4)^2} $$

So, the derivative of $ f(x) = \frac{x+1}{x-4} $ is $ f'(x) = \frac{-8}{(x-4)^2} $.