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)=(x)/(3x+1)`

Applying the limit definition

`f^(\prime)(x)=\lim _(h\to0)(f(x+h)-f(x))/(h)`

substitute given values

`\lim _(h\to0)((x+h)/(3(x+h)+1)-(x)/(3x+1))/(h)`
`\lim _(h\to0)((x+h)/(3x+3h+1)-(x)/(3x+1))/(h)`
`\lim _(h\to0)\text{ }(((3x+1)(x+h)-(3x+3h+1)x)/((3x+3h+1)(3x+1)))/(h)`

simplify

`\lim _(h\to0)\text{ }(((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{ }(1)/((9x^2+6x+9xh+3h+1))=(1)/((9x^2+6x+1))`

therefore

`f^(\prime)(x)=(1)/((9x^2+6x+1))`

simplify

`f^(\prime)(x)=(1)/((3x+1)^2)`