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

1 Answer
Jul 10, 2015

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.