How do you differentiate f(x)=#1/sqrt(x-4)# using first principles?

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Answer 1 · 2016-11-22T13:50:55

#f(x)=1/sqrt(x-4)#

#f'(x)=lim_(hrarr0) (f(x+h)-f(x))/h#

# = lim_(hrarr0) (1/sqrt((x-4)+h)-1/sqrt((x-4)))/h#

# = lim_(hrarr0) ((sqrt(x-4)-sqrt((x-4)+h))/(sqrt((x-4)+h)sqrt((x-4))))/(h/1)#

# = lim_(hrarr0) (sqrt(x-4)-sqrt((x-4)+h))/(sqrt((x-4)+h)sqrt((x-4))) * 1/h#

# = lim_(hrarr0) ((sqrt(x-4)-sqrt((x-4)+h)))/(hsqrt((x-4)+h)sqrt((x-4))) * ((sqrt(x-4)+sqrt((x-4)+h)))/((sqrt(x-4)+sqrt((x-4)+h)))#

# = lim_(hrarr0) ((x-4)-((x-4)+h))/(hsqrt((x-4)+h)sqrt((x-4))(sqrt(x-4)+sqrt((x-4)+h)))#

# = lim_(hrarr0) (-cancel(h)^1)/(cancel(h)sqrt((x-4)+h)sqrt((x-4))(sqrt(x-4)+sqrt((x-4)+h)))#

# = lim_(hrarr0) (-1)/(sqrt((x-4)+h)sqrt((x-4))(sqrt(x-4)+sqrt((x-4)+h)))#

# = (-1)/(sqrt((x-4)+0)sqrt((x-4))(sqrt(x-4)+sqrt((x-4)+0)))#

# = (-1)/((x-4)(2sqrt(x-4))) =(-1)/(2(x-4)sqrt(x-4)) = (-1)/(2(x-4)^(3/2))#