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!

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