Find the derivative of the function using the definition of derivative. g(t) = 3/ √t

Solution:

Given, g(t) = 3/√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) = [3/√(t + h) - (3/√t)]/h

g’(t) =\(\dfrac{3√t - 3√(t + h)}{h.(√t + h)(√t)}\)

g’(t) =\(\dfrac{[3(√t - √(t + h))]}{h.(√t + h)(√t)}\)

By using the conjugate of √t - √(t + h)

Multiplying by \(\dfrac{√t + √(t + h)}{√t + √(t + h)}\)

g’(t) = \(\lim_{h \to 0}\) \(\dfrac{3(√t - √(t+h)}{h.(√t+h)(√t)} . \dfrac{√t + √(t+h)}{√t + √(t+h)}\)

g’(t) = \(\lim_{h \to 0}\frac{3(t-(t+h))}{h(\sqrt{t+h}\sqrt{t})(\sqrt{t}+\sqrt{t+h})}\)

= \(\lim_{h \to 0}\frac{-3h}{h(\sqrt{t+h}\sqrt{t})(\sqrt{t}+\sqrt{t+h})}\)

= \(\lim_{h \to 0}\frac{-3}{(\sqrt{t+h}\sqrt{t})(\sqrt{t}+\sqrt{t+h})}\)

Now, evaluating the limit,

g’(t) = \(\frac{-3}{(\sqrt{t+0}\sqrt{t})(\sqrt{t}+\sqrt{t+0})}\)

g’(t) = \(\frac{-3}{(\sqrt{t}\sqrt{t})(\sqrt{t}+\sqrt{t})}\)

g’(t) = \(\frac{-3}{(t)(2\sqrt{t})}\)

g’(t) = \(\frac{-3}{2t\sqrt{t}}\)

Therefore, the derivative of the function is \(\frac{-3}{2t\sqrt{t}}\).

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Find the derivative of the function using the definition of derivative. g(t) = 3/ √t

Summary:

The derivative of the function using the definition of derivative. g(t) = 3/√ t is \(\frac{-3}{2t\sqrt{t}}\).