Answer :
The derivative of f(x) using first principles results in [tex]f'(x) = \frac{6}{(3x - 5)^2}[/tex]. The function's derivative is calculated explicitly.
The function given is [tex]f(x) = \frac{-2}{3x-5}[/tex]. To find its derivative using first principles, we'll use the definition of the derivative:
[tex]f'(x) = lim_{h \to 0} \frac{f(x + h) - f(x)}{h}[/tex]
Let's first compute f(x + h):
[tex]f(x + h) = \frac{-2}{3(x + h) - 5} = \frac{-2}{3x + 3h - 5}[/tex]
Now, we'll find the difference quotient:
[tex]\frac{f(x + h) - f(x)}{h} = \frac{\frac{-2}{3x+3h-5} - \frac{-2}{3x-5}}{h}[/tex]
To simplify this expression, find a common denominator:
[tex]\frac{f(x + h) - f(x)}{h} = \frac{-2 \cdot (3x - 5) - (-2) \cdot (3x + 3h - 5)}{h \cdot (3x + 3h - 5) \cdot (3x - 5)}[/tex]
Simplify the numerator:
[tex]-2(3x - 5) + 2(3x + 3h - 5) = -6x + 10 + 6x + 6h - 10 = 6h[/tex]
Now, simplify the entire fraction:
[tex]\frac{6h}{h \cdot (3x + 3h - 5) \cdot (3x - 5)} = \frac{6}{(3x + 3h - 5)(3x - 5)}[/tex]
As h approaches 0:
[tex]f'(x) = \lim_{h \to 0} \frac{6}{(3x + 3h - 5)(3x - 5)} = \frac{6}{(3x - 5)(3x - 5)} = \frac{6}{(3x - 5)^2}[/tex]
Therefore, the derivative of [tex]f(x) = \frac{-2}{3x-5}[/tex] is:
[tex]f'(x) = \frac{6}{(3x - 5)^2}[/tex]
