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

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Answer 1 · 2017-06-08T00:29:01

#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))#