Find the derivative of the function using the definition of the derivative. g(t) = 5/√t.

Solution:

Given, g(t) = 5/√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) = [5/√(t + h) - (5/√t)]/h

g’(t) = \(\dfrac{5√t - 5√(t + h)}{h.(√t + h)(√t)}\)

g’(t) = \(\dfrac{5(√t - √(t + h))}{h.(√t + h)(√t)}\)

= \(\dfrac{-5}{(√t + h)(√t)}\)

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

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

g’(t) = \(\lim_{h \to 0}\)\(\dfrac{-5}{.(√t + h)(√t)}\) . \(\dfrac{√t + √(t+h}{√t + √(t+h)}\)

= \(\lim_{h \to 0}\frac{-5}{(\sqrt{t+h}\sqrt{t})(\sqrt{t}+\sqrt{t+h})}\)

Now, evaluating the limit,

g’(t) = \(\frac{-5}{(\sqrt{t+0}\sqrt{t})(\sqrt{t}+\sqrt{t+0})}\)

g’(t) = \(\frac{-5}{(\sqrt{t}\sqrt{t})(\sqrt{t}+\sqrt{t})}\)

g’(t) = \(\frac{-5}{(t)(2\sqrt{t})}\)

g’(t) = \(\frac{-5}{2t\sqrt{t}}\)

Therefore, the derivative of the function is \(\frac{-5}{2t\sqrt{t}}\).

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Find the derivative of the function using the definition of derivative. g(t) = 5/√t.

Summary:

The derivative of the function using the definition of derivative. g(t) = 5/√ t is \(\frac{-5}{2t\sqrt{t}}\).