Question

How do you find the derivative of the function using the definition of derivative `g(t) = 7/sqrt(t)`?

Answer 1

The key step is to rationalize a numerator.

Explanation:

`g(t) = 7/sqrtt`

I'll assume that you are permitted to use the definition:

`g'(t) = lim_(hrarr0)(g(t+h)-g(t))/h`

(There are other ways of expressing the definition of derivative, but this is a very common one.)

`g'(t) = lim_(hrarr0)(g(t+h)-g(t))/h`

`= lim_(hrarr0)(7/sqrt(t+h)-7/sqrtt)/h`

`= lim_(hrarr0)(7sqrtt -7sqrt(t+h))/(sqrt(t+h)sqrtt)*1/h`

`= lim_(hrarr0)(7(sqrtt -sqrt(t+h)))/(hsqrt(t+h)sqrtt)`

Notice that, if we try to evaluate by substitution, we get the indeterminate form `0/0`.
The thing to try here (it will work) is to rationalize the numerator by using the conjugate of `sqrtt-sqrt(t+h)`.

That is: we will multiply by `1`, in the form: `(sqrtt + sqrt(t+h))/(sqrtt + sqrt(t+h))`

We resume:

`g'(t) = lim_(hrarr0)(7(sqrtt -sqrt(t+h)))/(hsqrt(t+h)sqrtt) *((sqrtt + sqrt(t+h)))/((sqrtt + sqrt(t+h)))`

`=lim_(hrarr0) (7(t-(t+h)))/(hsqrt(t+h)sqrtt(sqrtt + sqrt(t+h))`

`=lim_(hrarr0) (-7cancel(h))/(cancel(h)sqrt(t+h)sqrtt(sqrtt + sqrt(t+h))`

Now we can evaluate the limit:

`g'(t) = (-7)/(sqrt(t+0)sqrtt(sqrtt + sqrt(t+0))`

`= (-7)/(sqrttsqrtt(2sqrtt)) = (-7)/(t(2sqrtt)) = (-7)/(2tsqrtt)`

Note
It may be helpful to observe that in some sense we have traded the subtraction: `sqrtt-sqrt(t+h)` in the numerator for an addition: `sqrtt+sqrt(t+h)` in the denominator.
The subtraction goes to `0`, the addition does not.
In the process, we were able to eliminate the factor of `h` from both the numerator and denominator.