Question

How do you use the limit definition of the derivative to find the derivative of `f(x)=sqrt(x+8)+3x`?

Answer 1

`f'(x) = 1/(2sqrt(x+8)) +3`

Explanation:

The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So with `f(x) = sqrt(x+8)+3x` then;

`\ \ \ \ \ f(x+h) = sqrt((x+h)+8)+3(x+h)`
`:. f(x+h) = sqrt(x+h+8)+3x+3h`

So Then the numerator of the derivative is:
`f(x+h)-f(x)=(sqrt(x+h+8)+3x+3h) - (sqrt(x+8)+3x)`
`" "=sqrt(x+h+8)+3x+3h - sqrt(x+8)-3x`
`" "=sqrt(x+h+8) - sqrt(x+8) +3h`
`" "=(sqrt(x+h+8) - sqrt(x+8))*(sqrt(x+h+8) + sqrt(x+8))/(sqrt(x+h+8) + sqrt(x+8)) +3h`

`" "=(sqrt(x+h+8)^2 - sqrt(x+8)^2)/(sqrt(x+h+8) + sqrt(x+8)) +3h`
`" "=(x+h+8 - (x+8))/(sqrt(x+h+8) + sqrt(x+8)) +3h`
`" "=(x+h+8 - x-8)/(sqrt(x+h+8) + sqrt(x+8)) +3h`
`" "=(h)/(sqrt(x+h+8) + sqrt(x+8)) +3h`

And so the derivative of `y=f(x)` is given by:

`\ \ \ \ \ f'(x) = lim_(h rarr 0) ( f(x+h)-f(x) ) / h`
`" " = lim_(h rarr 0) ( (h)/(sqrt(x+h+8) + sqrt(x+8)) +3h) / h`
`" " = lim_(h rarr 0) ( 1/(sqrt(x+h+8) + sqrt(x+8)) +3)`
`" " = 1/(sqrt(x+8) + sqrt(x+8)) +3`
`:. f'(x) = 1/(2sqrt(x+8)) +3`