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

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Answer 1 · 2018-03-31T22:11:38

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

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)#