How do you use the limit definition to find the derivative of #f(x)=sqrt(3-2x)#?

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Answer 1 · 2017-01-03T17:51:13

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

#f'(x) = lim_(h->0) (sqrt(3 - 2(x + h)) - sqrt(3 - 2x))/h#

#f'(x) = lim_(h->0) (sqrt(3 - 2x - 2h) - sqrt(3 - 2x))/h#

Multiply the entire expression by the conjugate of the numerator, which is #sqrt(3 - 2x - 2h) + sqrt(3 - 2x)#.

#f'(x) = lim_(h->0) (sqrt(3 - 2x - 2h) - sqrt(3 - 2x))/h * (sqrt(3 - 2x - 2h) + sqrt(3 - 2x))/(sqrt(3 - 2x - 2h) + sqrt(3 - 2x))#

This creates a difference of squares and gets rid of the #√#'s in the numerator.

#f'(x) = lim_(h->0) (3 - 2x - 2h- (3 - 2x))/(h(sqrt(3 - 2x - 2h) + sqrt(3 - 2x))#

#f'(x) = lim_(h->0) (-2h)/(h(sqrt(3- 2x - 2h) + sqrt(3 - 2x))#

#f'(x) = lim_(h-> 0) -2/(sqrt(3 - 2x - 2h) + sqrt(3 - 2x)#

You can now use substitution to evaluate.

#f'(x) = -2/(sqrt(3 - 2x - 2(0)) + sqrt(3 - 2x))#

#f'(x) = -2/(sqrt(3 - 2x) + sqrt(3 - 2x)#

#f'(x) = -2/(2sqrt(3 - 2x)#

#f'(x) = -1/sqrt(3 - 2x)#

If you were to check this using the chain rule, you would get the same result.

Hopefully this helps!