Using the limit definition, how do you find the derivative of #f(x) = -5x^2+8x+2#?

2 Answers
Jun 17, 2016

#f'(x)= -10x+8#

Explanation:

Using the limit definition to find the derivativeof #f(x)#
#f'(x)=lim_(h->0)(f(x+h)-f(x))/h#
so it can be expressed as following
#lim_(h->0)((-5(x+h)^2+8(x+h)+2)-(-5x^2+8x+2))/h#
=#-10x+8#

Jun 17, 2016

Use the formula #f'(x) = lim_(h-> 0) (f(x + h) - f(x))/h#

Explanation:

#f'(x) = lim_(h->0) (-5(x + h)^2 + 8(x + h) + 2 -(-5x^2 + 8x + 2))/h#

#f'(x) = lim_(h->0)(-5(x^2 + 2xh + h^2) + 8x + 8h + 2 + 5x^2 - 8x - 2)/h#

#f'(x)= lim_(h->0) (-5x^2 - 10xh - 5h^2 + 8x + 8h + 2 + 5x^2 - 8x - 2)/h#

#f'(x) = lim_(h->0) (-10xh - 5h^2+ 8h)/h#

#f'(x) = lim_(h->0) (cancel(h)(-10x - 5h + 8))/cancel(h)#

#f'(x) = -10x - 5(0) + 8#

#f'(x) = -10x + 8#

#:.# The derivative is #dy/dx = -10x + 8#.

Hopefully this helps!