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

Question

Luke Alvarez · Accepted Answer

The key step is to rationalize a numerator.

#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.

Aurora Alvarado · Answer

undefined To find the derivative of the function $ g(t) = \frac{7}{\sqrt{t}} $ using the definition of derivative:

1. Start with the definition of the derivative: 
$$ g'(t) = \lim_{h \to 0} \frac{g(t + h) - g(t)}{h} $$

2. Substitute the function $ g(t) = \frac{7}{\sqrt{t}} $ into the definition:
$$ g'(t) = \lim_{h \to 0} \frac{\frac{7}{\sqrt{t + h}} - \frac{7}{\sqrt{t}}}{h} $$

3. Rationalize the numerator:
$$ g'(t) = \lim_{h \to 0} \frac{7\sqrt{t} - 7\sqrt{t + h}}{h\sqrt{t}\sqrt{t + h}} $$

4. Combine like terms in the numerator:
$$ g'(t) = \lim_{h \to 0} \frac{7(\sqrt{t} - \sqrt{t + h})}{h\sqrt{t}\sqrt{t + h}} $$

5. Multiply and divide by the conjugate of the numerator:
$$ g'(t) = \lim_{h \to 0} \frac{7(\sqrt{t} - \sqrt{t + h})}{h(\sqrt{t} - \sqrt{t + h})(\sqrt{t} + \sqrt{t + h})} $$

6. Simplify the expression:
$$ g'(t) = \lim_{h \to 0} \frac{7}{h(\sqrt{t} + \sqrt{t + h})} $$

7. Factor out common terms:
$$ g'(t) = \lim_{h \to 0} \frac{7}{h} \cdot \frac{1}{\sqrt{t} + \sqrt{t + h}} $$

8. Take the limit as $ h $ approaches 0:
$$ g'(t) = \frac{7}{2t\sqrt{t}} $$

Therefore, the derivative of the function $ g(t) = \frac{7}{\sqrt{t}} $ with respect to $ t $ is $ g'(t) = \frac{7}{2t\sqrt{t}} $.