# What is the instantaneous rate of change for #f(t)= sqrt (5t+1)# when t=1?

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To find the instantaneous rate of change for ( f(t) = \sqrt{5t + 1} ) when ( t = 1 ), we need to find the derivative of ( f(t) ) with respect to ( t ) and then evaluate it at ( t = 1 ).

First, we find the derivative:

[ f'(t) = \frac{d}{dt}\left(\sqrt{5t + 1}\right) = \frac{1}{2\sqrt{5t + 1}} \cdot \frac{d}{dt}(5t + 1) = \frac{5}{2\sqrt{5t + 1}} ]

Now, we evaluate the derivative at ( t = 1 ):

[ f'(1) = \frac{5}{2\sqrt{5(1) + 1}} = \frac{5}{2\sqrt{6}} = \frac{5\sqrt{6}}{12} ]

Therefore, the instantaneous rate of change for ( f(t) = \sqrt{5t + 1} ) when ( t = 1 ) is ( \frac{5\sqrt{6}}{12} ).

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