How do you find the derivative using the difference quotient #f(x) = 5 / (x^2 + 7)#?

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Answer 1 · 2015-04-12T10:22:00

#y=f(x)/g(x)rArry'=(f'(x)g(x)-f(x)g'(x))/[g(x)]^2#.

So:

#y'=(0*(x^2+7)-5*2x)/(x^2+7)^2=(-10x)/(x^2+7)^2#.

Answer 2 · 2015-04-12T16:03:58

The details depend on which difference quotient you use :

#f'(a) = lim_(xrarra) (f(x) - f(a))/(x-a)#

or,

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

For #f(x) = 5/(x^2+7)# and using the second form:

#f'(x) = lim_(hrarr0) ( 5/((x+h)^2+7) - 5/(x^2+7))/(h)#.

#= lim_(hrarr0) ( 5(x^2+7) - 5 ((x+h)^2+7))/(h((x+h)^2+7)(x^2+7))#

#= lim_(hrarr0) ( 5x^2+35 - 5x^2-10xh-5h^2-35)/(h((x+h)^2+7)(x^2+7))#

#= lim_(hrarr0) ( -10x-5h)/(((x+h)^2+7)(x^2+7))#

#= ( -10x)/((x^2+7)^2)#