How do you find f'(x) using the limit definition given #f(x)=4/sqrt(x-5)#?

Question

William Abdalla · Accepted Answer

#d/dx 4/sqrt(x-5) = -2/((x-5) sqrt(x-5))#

By definition the derivative of a function #f(x)# is:

#(df)/dx = lim_(h->0) (f(x+h)-f(x))/h#

then we have:

#d/dx 4/sqrt(x-5) = lim_(h->0) (4/sqrt(x-5+h)-4/sqrt(x-5))/h = 4lim_(h->0)1/h(1/sqrt(x-5+h)-1/sqrt(x-5))#

#d/dx 4/sqrt(x-5) = 4lim_(h->0)1/h((sqrt(x-5)-sqrt(x-5+h))/(sqrt(x-5+h)sqrt(x-5)))#

Multiply and divide by: #(sqrt(x-5)+sqrt(x-5+h))# and keep in mind that: #(a+b)(a-b) = (a^2-b^2)#

#d/dx 4/sqrt(x-5) = 4lim_(h->0)1/h((sqrt(x-5)-sqrt(x-5+h))/(sqrt(x-5+h)sqrt(x-5))) xx (sqrt(x-5+h)+sqrt(x-5))/(sqrt(x-5+h)+sqrt(x-5))#

#d/dx 4/sqrt(x-5) = 4lim_(h->0)1/h(((x-5)-(x-5+h))/((sqrt(x-5+h)sqrt(x-5)) (sqrt(x-5+h)+sqrt(x-5))))#

#d/dx 4/sqrt(x-5) = -4lim_(h->0)1/h((h)/((sqrt(x-5+h)sqrt(x-5)) (sqrt(x-5+h)+sqrt(x-5))))#

#d/dx 4/sqrt(x-5) = -4lim_(h->0)(1/((sqrt(x-5+h)sqrt(x-5)) (sqrt(x-5+h)+sqrt(x-5))))#

#d/dx 4/sqrt(x-5) = -4/(2(x-5) sqrt(x-5))#

#d/dx 4/sqrt(x-5) = -2/((x-5) sqrt(x-5))#

Luke Alvarez · Answer

undefined To find $f'(x)$ using the limit definition, you use the formula:

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

Given $f(x) = \frac{4}{\sqrt{x - 5}}$, substitute into the formula:

$$f'(x) = \lim_{h \to 0} \frac{\frac{4}{\sqrt{x + h - 5}} - \frac{4}{\sqrt{x - 5}}}{h}$$

Now, simplify the expression and evaluate the limit.