Using the limit definition, how do you find the derivative of #f(x) = x/(x+4)#?

Question

Evelyn Agard · Accepted Answer

# = 4 / ((x+4)^2)# #f(x) = x/(x+4)# #f'(x)=lim_{h to 0} (f(x+h) - f(x))/(h)# # =lim_{h to 0} 1/(h) ((x+h)/(x + h +4) - x/(x+4))# # =lim_{h to 0} 1/(h) ( (x+h)(x+4) - x(x+h+4))/ ((x+h+4)(x+4))# # =lim_{h to 0} 1/(h) ( (x^2 + 4x + hx + 4h - x^2- xh-4x))/ ((x+h+4)(x+4))# # =lim_{h to 0} 1/(h) ( 4h )/ ((x+h+4)(x+4))# # =lim_{h to 0} ( 4 )/ ((x+h+4)(x+4))# # = ( 4 )/ ((x+4)(x+4))# # = 4 / ((x+4)^2)#

Paisley Johnson · Answer

undefined To find the derivative of $ f(x) = \frac{x}{x+4} $ using the limit definition of a derivative, we 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 function $ f(x) = \frac{x}{x+4} $ into the definition:

$$ f'(x) = \lim_{h \to 0} \frac{\frac{x+h}{x+h+4} - \frac{x}{x+4}}{h} $$

3. Simplify the expression:

$$ f'(x) = \lim_{h \to 0} \frac{(x+h)(x+4) - x(x+h+4)}{h(x+4)(x+h+4)} $$

4. Expand the numerator:

$$ f'(x) = \lim_{h \to 0} \frac{x^2 + 4x + hx + 4h - x^2 - hx - 4x}{h(x+4)(x+h+4)} $$

5. Cancel out like terms:

$$ f'(x) = \lim_{h \to 0} \frac{4h}{h(x+4)(x+h+4)} $$

6. Simplify further:

$$ f'(x) = \lim_{h \to 0} \frac{4}{(x+4)(x+h+4)} $$

7. Evaluate the limit:

$$ f'(x) = \frac{4}{(x+4)(x+4)} $$

$$ f'(x) = \frac{4}{(x+4)^2} $$

So, the derivative of $ f(x) = \frac{x}{x+4} $ with respect to $ x $ is $ f'(x) = \frac{4}{(x+4)^2} $.