How do you use the limit definition to find the derivative of #f(x)=1/(4x-3)#?

Question

Ezra Adams · Accepted Answer

#f'(x) = lim_(Deltax->0) (Deltaf)/(Deltax) =(-4)/((4x-3)^2)#

The definition of the derivative of #f(x)# is:

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

Let us calculate the increment of #f(x)#:

#Deltaf = 1/(4(x+Deltax)-3) -1/(4x-3)#

#Deltaf = (4x-3-4x-4Deltax+3)/((4x+4Deltax-3)(4x-3))=(-4Deltax)/((4x+4Deltax-3)(4x-3))#

So:

#(Deltaf)/(Deltax) =(-4)/((4x+4Deltax-3)(4x-3))#

and

#lim_(Deltax->0) (Deltaf)/(Deltax) =(-4)/((4x-3)^2)#

Axel Alvarez · Answer

undefined To use the limit definition to find the derivative of $ f(x) = \frac{1}{4x - 3} $:

1. Start with the definition of the derivative:
$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

2. Substitute the function $ f(x) = \frac{1}{4x - 3} $ into the formula:
$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{4(x + h) - 3} - \frac{1}{4x - 3}}{h} $$

3. Simplify the expression:
$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{4x + 4h - 3} - \frac{1}{4x - 3}}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{(4x - 3) - (4x + 4h - 3)}{h(4x + 4h - 3)(4x - 3)} $$
$$ f'(x) = \lim_{h \to 0} \frac{4x - 3 - 4x - 4h + 3}{h(4x + 4h - 3)(4x - 3)} $$
$$ f'(x) = \lim_{h \to 0} \frac{-4h}{h(4x + 4h - 3)(4x - 3)} $$

4. Cancel out the $ h $ in the numerator and denominator:
$$ f'(x) = \lim_{h \to 0} \frac{-4}{(4x + 4h - 3)(4x - 3)} $$

5. Evaluate the limit as $ h $ approaches 0:
$$ f'(x) = \frac{-4}{(4x - 3)(4x - 3)} $$

6. Simplify the expression:
$$ f'(x) = \frac{-4}{(4x - 3)^2} $$

Thus, the derivative of $ f(x) = \frac{1}{4x - 3} $ is $ f'(x) = \frac{-4}{(4x - 3)^2} $.