How do you find f'(x) using the limit definition given #f (x) = sqrt(1+3x)#?

Question

Emily Ammerman · Accepted Answer

#= 3 / (2sqrt(1+3x )# #f (x) = sqrt(1+3x)# by definition #f'(x) = lim_{h to 0} (f(x+h) - f(x))/h# #= lim_[h to 0] 1/h (sqrt(1+3(x+h)) - sqrt(1+3x))# multiply by conjugate #= lim_[h to 0] 1/h (sqrt(1+3(x+h)) - sqrt(1+3x)) times (sqrt(1+3(x+h)) + sqrt(1+3x))/(sqrt(1+3(x+h)) + sqrt(1+3x))# #= lim_[h to 0] 1/h (1+3(x+h) - (1+3x)) / (sqrt(1+3(x+h)) + sqrt(1+3x))# #= lim_[h to 0] 1/h (3h) / (sqrt(1+3(x+h)) + sqrt(1+3x))# #= lim_[h to 0] (3) / (sqrt(1+3(x+h)) + sqrt(1+3x))# #= (3) / (sqrt(1+3(x)) + sqrt(1+3x))# #= 3 / (2sqrt(1+3x )#

Samantha Adkison · Answer

undefined To find $ f'(x) $ using the limit definition for $ f(x) = \sqrt{1 + 3x} $, you apply the definition of the derivative:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substitute $ f(x) = \sqrt{1 + 3x} $ into the formula:

$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{1 + 3(x + h)} - \sqrt{1 + 3x}}{h} $$

Multiply the numerator and denominator by the conjugate of the numerator to eliminate the radical:

$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{1 + 3(x + h)} - \sqrt{1 + 3x}}{h} \cdot \frac{\sqrt{1 + 3(x + h)} + \sqrt{1 + 3x}}{\sqrt{1 + 3(x + h)} + \sqrt{1 + 3x}} $$

Simplify the numerator:

$$ f'(x) = \lim_{h \to 0} \frac{(1 + 3(x + h)) - (1 + 3x)}{h(\sqrt{1 + 3(x + h)} + \sqrt{1 + 3x})} $$

$$ f'(x) = \lim_{h \to 0} \frac{3h}{h(\sqrt{1 + 3(x + h)} + \sqrt{1 + 3x})} $$

$$ f'(x) = \lim_{h \to 0} \frac{3}{\sqrt{1 + 3(x + h)} + \sqrt{1 + 3x}} $$

Now, substitute $ h = 0 $ into the expression:

$$ f'(x) = \frac{3}{\sqrt{1 + 3x} + \sqrt{1 + 3x}} $$

$$ f'(x) = \frac{3}{2\sqrt{1 + 3x}} $$