Question

How do you determine the instantaneous rate of change of `y(x) = sqrt(3x + 1)` for `x = 1`?

Answer 1

Apply the chain rule for derivatives and substitute `x=1` in the result.

`(d (y(x)))/(dx) = (1/2)(3x+1)^(-1/2)*(3)`

`(d (y(1)))/(dx) = 3/(2 sqrt(3+1)) = 3/4`

`(d (y(1)))/(dx)` is the instantaneous rate of change of `y(x)` at `x=1`

Answer 2

If you need to do this from the definition, the details will depend on your textbook and instructor's choice of definition:

You'll have one of:

`lim_(xrarr1)(y(x)-y(1))/(x-1)` or `lim_(hrarr0)(y(1+h)-y(1))/h`

Definition 1:

`lim_(xrarr1)(y(x)-y(1))/(x-1)=lim_(xrarr1)(sqrt(3x+1)-sqrt(3(1)+1))/(x-1)`

`=lim_(xrarr1)(sqrt(3x+1)-2)/(x-1)`

Notice that substitution leads to indeterminate form: `0/0`. The trick (technique) to try in this case is to rationalize the numerator. Multiply numerator and denominator by the conjugate of the numerator. (Multiply the top and bottom by `(sqrt(3x+1) + 2)`

`lim_(xrarr1)(y(x)-y(1))/(x-1)=lim_(xrarr1)[((sqrt(3x+1)-2))/(x-1) ((sqrt(3x+1) + 2))/((sqrt(3x+1) + 2))]`

`=lim_(xrarr1)(sqrt(3x+1)^2-2^2)/((x-1)(sqrt(3x+1) + 2)`

`=lim_(xrarr1)(3x+1-4)/((x-1)(sqrt(3x+1) + 2)`

`=lim_(xrarr1)(3x-3)/((x-1)(sqrt(3x+1) + 2)`

`=lim_(xrarr1)(3(x-1))/((x-1)(sqrt(3x+1) + 2)` (Still form `0/0`)

`=lim_(xrarr1)3/(sqrt(3x+1) + 2)` (No longer form `0/0`)

`=3/(sqrt(3(1)+1) + 2)=3/(2+2)=3/4`

Definition 2:

`lim_(hrarr0)(y(1+h)-y(1))/h =lim_(hrarr0)(sqrt(3(1+h)+1)-sqrt(3(1)+1))/h`

`=lim_(hrarr0)(sqrt(3h+4)-2)/h`

Notice that substitution leads to indeterminate form: `0/0`. The trick (technique) to try in this case is to rationalize the numerator. Multiply numerator and denominator by the conjugate of the numerator. (Multiply the top and bottom by `(sqrt(3h+4) + 2)`

`lim_(hrarr0)(y(1+h)-y(1))/h =lim_(hrarr0)((sqrt(3h+4)-2))/h ((sqrt(3h+4) + 2))/((sqrt(3h+4) + 2))`

`=lim_(hrarr0)((sqrt(3h+4)^2-2^2))/(h(sqrt(3h+4) + 2))`

`=lim_(hrarr0)(3h+4-4)/(h(sqrt(3h+4) + 2))` , (still form `0/0`)

`=lim_(hrarr0)(3h)/(h(sqrt(h+4) + 2))` , (still form `0/0`)

`=lim_(hrarr0)3/(sqrt(h+4) + 2)` , (no longer form `0/0`)

`=3/(sqrt(0+4) + 2)=3/(2+2)=3/4`