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
1)
2)
Explanation:
1)
According to the limit definition, the derivative of
#f'(x) = lim_(h-> 0) (f(x+h) - f(x))/h#
To compute
#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
# = lim_(h-> 0) (h(2x + h - 6))/h#
... cancel
# = lim_(h-> 0) (2x + h - 6)#
... apply the limit, so plug
# = 2x - 6#
So, we have found that
#f'(x) = 2x -6# .
================================
2)
#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
#= 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
#= 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
#= lim_(h->0) (4) / (sqrt(x+h) + sqrt(x))#
... apply the limit, so plug
#= 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)#
If
If
Explanation:
Question 1:
(using limit definition for derivative)
Question 2:
(using limit definition for derivative)