How do you use the definition of a derivative to find the derivative of f(x) = 4 / sqrt( 5 - x )?

1 Answer
Mar 31, 2018

d/dx( 4/sqrt(5-x)) =2/((5-x)^(3/2)

Explanation:

Using the definition of derivative:

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

we have:

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

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

Rationalize the numerator by multiplying and dividing by:
(sqrt(5-x) + sqrt(5-x-h)) and using the algebraic identity:
(a-b)(a+b) = a^2-b^2

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

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

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

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

d/dx( 4/sqrt(5-x)) = 4/(sqrt(5-x) sqrt(5-x)(sqrt(5-x) + sqrt(5-x))

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

d/dx( 4/sqrt(5-x)) =2/((5-x)^(3/2)