Question

For function a, find the general derivative as a new function by using the limit definition

Answer 1

we have the function

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

Applying the limit definition

$f^{\prime}(x)=\lim _{h\to0}\frac{f(x+h)-f(x)}{h}$

substitute given values

$\lim _{h\to0}\frac{\frac{x+h}{3(x+h)+1}-\frac{x}{3x+1}}{h}$$\lim _{h\to0}\frac{\frac{x+h}{3x+3h+1}-\frac{x}{3x+1}}{h}$$\lim _{h\to0}\text{ }\frac{\frac{(3x+1)(x+h)-(3x+3h+1)x}{(3x+3h+1)(3x+1)}}{h}$

simplify

$\lim _{h\to0}\text{ }\frac{\frac{(3x^2+3xh+x+h)-(3x^2+3hx+x)}{(9x^2+3x+9xh+3h+3x+1)}}{h}$$\lim _{h\to0}\text{ }\frac{\frac{h^{}}{(9x^2+6x+9xh+3h+1)}}{h}$$\lim _{h\to0}\text{ }\frac{1}{(9x^2+6x+9xh+3h+1)}=\frac{1}{(9x^2+6x+1)}$

therefore

$f^{\prime}(x)=\frac{1}{(9x^2+6x+1)}$

simplify

$f^{\prime}(x)=\frac{1}{(3x+1)^2}$