# How do you use the definition of a derivative to find the derivative of #g(t) = 7/sqrt(t)#?

Now, by generalised Binomial Theorem:

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To find the derivative of ( g(t) = \frac{7}{\sqrt{t}} ), we use the definition of a derivative, which states that the derivative of a function represents the rate of change of that function with respect to its independent variable.

Using the definition of the derivative, we have:

[ g'(t) = \lim_{{h \to 0}} \frac{g(t + h) - g(t)}{h} ]

Substituting ( g(t) = \frac{7}{\sqrt{t}} ), we get:

[ g'(t) = \lim_{{h \to 0}} \frac{\frac{7}{\sqrt{t+h}} - \frac{7}{\sqrt{t}}}{h} ]

Now, we'll simplify the expression:

[ g'(t) = \lim_{{h \to 0}} \frac{7\sqrt{t} - 7\sqrt{t+h}}{h\sqrt{t}\sqrt{t+h}} ]

[ g'(t) = \lim_{{h \to 0}} \frac{7\sqrt{t} - 7\sqrt{t+h}}{h\sqrt{t}\sqrt{t+h}} \times \frac{\sqrt{t} + \sqrt{t+h}}{\sqrt{t} + \sqrt{t+h}} ]

[ g'(t) = \lim_{{h \to 0}} \frac{7t - 7(t+h)}{h\sqrt{t}\sqrt{t+h}(\sqrt{t} + \sqrt{t+h})} ]

[ g'(t) = \lim_{{h \to 0}} \frac{7t - 7t - 7h}{h\sqrt{t}\sqrt{t+h}(\sqrt{t} + \sqrt{t+h})} ]

[ g'(t) = \lim_{{h \to 0}} \frac{-7h}{h\sqrt{t}\sqrt{t+h}(\sqrt{t} + \sqrt{t+h})} ]

[ g'(t) = \lim_{{h \to 0}} \frac{-7}{\sqrt{t}\sqrt{t+h}(\sqrt{t} + \sqrt{t+h})} ]

Now, plugging in ( h = 0 ), we get:

[ g'(t) = \frac{-7}{2t^{3/2}} ]

Therefore, the derivative of ( g(t) = \frac{7}{\sqrt{t}} ) is ( g'(t) = \frac{-7}{2t^{3/2}} ).

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