How do you use the limit definition of the derivative to find the derivative of #f(x)=(x+1)/(x-4)#?

1 Answer
sjc
Dec 3, 2016

#f'(x)=(-5)/((x-4)^2)#

Explanation:

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

#f'(x)=Lim_(hrarr0)((((x+h+1)/(x+h-4))-((x+1)/(x-4)))/h)#

#f'(x)=Lim_(hrarr0)((((x+h+1)(x-4)-(x+h-4)(x+1))/((x+h-4)(x-4)))/h)#

#f'(x)=Lim_(hrarr0)(((x+h+1)(x-4)-(x+h-4)(x+1))/(h(x+h-4)(x-4)))#

#f'(x)=Lim_(hrarr0)((x^2-4x+hx-4h+x-4)-(x^2+x+hx+h-4x-4))/(h(x^2-4x+hx-4h-4x+16))#

#f'(x)=Lim_(hrarr0)(x^2-4x+hx-4h+x-4-x^2-x-hx-h+4x+4)/(h(x^2-4x+hx-4h+x-4))#

#f'(x)=Lim_(hrarr0)(cancel(x^2)-cancel(4x)+cancel(hx)-4h+cancel(x)-cancel(4)-cancel(x^2)-cancel(x)-cancel(hx)-h+cancel(4x)+cancel(4))/(h(x^2-4x+hx+4h-4x+16))#

#f'(x)=Lim_(hrarr0)(-5cancel(h))/(cancel(h)(x^2-4x+hx+4h-4x+16))#

#f'(x)=Lim_(hrarr0)(-5)/(x^2-4x+hx+4h-4x+16)#

in the limit #" "hx," "4h" "=0#

#f'(x)=(-5)/(x^2-4x-4x+16)#

#f'(x)=(-5)/(x^2-8x+16)#

#f'(x)=(-5)/((x-4)^2)#

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