Find the derivative of the function using the definition of derivative. g(t) = 9/√ t.

Solution:

Given, g(t) = 9/√t

We have to find the derivative of the function.

The differentiation by the first principle states that,

f’(x) = [f(x+h)-f(x)]/h 

Where h<<0

Now, g’(t) = [9/√(t+h) - (9/√t)]/h

g’(t) = [9√t - 9√(t+h)]/[h.(√t+h)(√t)]

g’(t) = [9(√t - √(t+h))]/[h.(√t+h)(√t)]

By using the conjugate of √t - √(t+h)

Multiplying by √t + √(t+h)/√t + √(t+h)

g’(t) = \(\lim_{h \to 0}\) [9(√t - √(t+h))]/[h.(√t+h)(√t)][√t + √(t+h)]/[√t + √(t+h)]

g’(t) = \(\lim_{h \to 0}\frac{9(t-(t+h))}{h(\sqrt{t+h}\sqrt{t})(\sqrt{t}+\sqrt{t+h})}\)

= \(\lim_{h \to 0}\frac{-9h}{h(\sqrt{t+h}\sqrt{t})(\sqrt{t}+\sqrt{t+h})}\)

= \(\lim_{h \to 0}\frac{-9}{(\sqrt{t+h}\sqrt{t})(\sqrt{t}+\sqrt{t+h})}\)

Now, evaluating the limit,

g’(t) = \(\frac{-9}{(\sqrt{t+0}\sqrt{t})(\sqrt{t}+\sqrt{t+0})}\)

g’(t) = \(\frac{-9}{(\sqrt{t}\sqrt{t})(\sqrt{t}+\sqrt{t})}\)

g’(t) = \(\frac{-9}{(t)(2\sqrt{t})}\)

g’(t) = \(\frac{-9}{2t\sqrt{t}}\)Therefore, the derivative of the function is \(\frac{-9}{2t\sqrt{t}}\).

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Find the derivative of the function using the definition of derivative. g(t) = 9/√ t.

Summary:

The derivative of the function using the definition of derivative. g(t) = 9/√ t is \(\frac{-9}{2t\sqrt{t}}\).