Using the limit definition for derivatives, how do you find the derivative of (a) #f(x)=x^2-6x#; (b) #f(x)=4sqrt(x)#?

2 Answers
Lotusbluete
Feb 12, 2016

1) #f'(x) = 2x - 6#

2) #f'(x) = 2/sqrt(x)#

Explanation:

1) # f(x) = x^2 - 6x#

According to the limit definition, the derivative of #f(x)# is

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

To compute #f(x+h)#, plug #x+h# for every occurance of #x# in #f(x)#:

#f(x+h) = (x+h)^2 - 6(x+h)#

Thus, you can compute your derivative as follows:

#f'(x) = lim_(h-> 0) ((x+h)^2 - 6(x+h) - (x^2 - 6x))/h#

# = lim_(h-> 0) (x^2 + 2xh + h^2 - 6x - 6h - x^2 + 6x)/h#

# = lim_(h-> 0) (color(blue)(cancel(x^2)) + 2xh + h^2 - color(red)(cancel(6x)) - 6h - color(blue)(cancel(x^2)) + color(red)(cancel(6x)))/h#

# = lim_(h-> 0) (2xh + h^2 - 6h)/h#

... factor #h# in the numerator...

# = lim_(h-> 0) (h(2x + h - 6))/h#

... cancel #h#...

# = lim_(h-> 0) (2x + h - 6)#

... apply the limit, so plug #h=0#...

# = 2x - 6#

So, we have found that

#f'(x) = 2x -6#.

================================

2) #f(x) = 4 sqrt(x)#

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

#= lim_(h-> 0) (4sqrt(x+h) - 4sqrt(x))/h#

#= lim_(h-> 0) (4(sqrt(x+h) - sqrt(x)))/h#

... multiply the numerator and the denominator with #(sqrt(x+h) + sqrt(x))#...

#= lim_(h-> 0) (4(sqrt(x+h) - sqrt(x))color(blue)((sqrt(x+h) + sqrt(x))))/(h color(blue)((sqrt(x+h) + sqrt(x))))#

... use the formula #(a+b)(a-b) = a^2 - b^2# to expand the numerator...

#= lim_(h->0) (4((sqrt(x+h))^2 - (sqrt(x))^2 )) / (h (sqrt(x+h) + sqrt(x)))#

#= lim_(h->0) (4(x + h - x)) / (h (sqrt(x+h) + sqrt(x)))#

#= lim_(h->0) (4h) / (h (sqrt(x+h) + sqrt(x)))#

... cancel #h#....

#= lim_(h->0) (4) / (sqrt(x+h) + sqrt(x))#

... apply the limit, so plug #h=0#...

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

#= 4 / (2 sqrt(x))#

#= 2 / sqrt(x)#

So you have found your derivative and it's

#f'(x) = 2 / sqrt(x)#

Alan P.
Feb 12, 2016

If #y=x^2-6x# then #dy/dx= 2x-6#

If #f(x)=4sqrt(x)# then #f'(x)=2/sqrt(x)#

Explanation:

Question 1: #color(black)(y=x^2-6x)#
(using limit definition for derivative)
#dy/dx = lim_(hrarr0)(((x+h)^2-6(x+h)) - (x^2-6x))/h#

#color(white)("XXX")=lim_(hrarro)(cancel(x^2)+2hx+h^2cancel(-6x)-6h-cancel(-x^2)cancel(+6x))/h#

#color(white)("XXX")=lim_(hrarr0)(2cancel(h)x-6cancel(h))/cancel(h)#

#color(white)("XXX")=2x-6#

Question 2: #color(black)(f(x)=4sqrt(x))#
(using limit definition for derivative)
#f'(x)=lim_(hrarr0)=(4sqrt(x+h)-4sqrt(x)/h#

#color(white)("XXX")=4lim_(hrarr0)((sqrt(x+h)-sqrt(x))/h)*((sqrt(x+h)+sqrt(x))/(sqrt(x+h)+sqrt(x)))#

#color(white)("XXX")=4lim_(hrarr0)(x+h-x)/(hsqrt(x+h)+sqrt(x))#

#color(white)("XXX")=4lim_(hrarr0)(cancel(h))/(cancel(h)sqrt(x+h)+sqrt(x))#

#color(white)("XXX")=4*1/(2sqrt(x))#

#color(white)("XXX")=2/sqrt(x)#

Related questions
Impact of this question
2036 views around the world
You can reuse this answer
Creative Commons License