Find the derivative of the function using the definition of derivative. g(x) = √8 - x.

Solution:

Given, g(x) = √(8 - x)

We have to find the derivative of the function.

Derivative by the first principle is stated as

f’(x) = [f(x + h) - f(x)]/h where h << 0

f(x + h) = √(8 - x + h)

So, g’(x) = [ √(8 + h - x) - √(8 - x)]/h

Now, multiplying and dividing by [ √(8 + h - x) + √(8 - x)] we get,

g’(x)=\(\frac{\sqrt{8+h-x}-\sqrt{8-x}}{h}\times\frac{\sqrt{8+h-x}+\sqrt{8-x}}{\sqrt{8+h-x}+\sqrt{8-x}}\)

We know, (a + b)(a - b) = a² - b²

So, g’(x) = \(\frac{(\sqrt{8+h-x})^{2}-(\sqrt{8-x})^{2}}{h(\sqrt{8+h-x}+\sqrt{8-x})}\)

g’(x) = \(\frac{(8+h-x)-(8-x)}{h(\sqrt{8+h-x}+\sqrt{8-x})}\)

g’(x) = \(\frac{h}{h(\sqrt{8+h-x}+\sqrt{8-x})}\)

g’(x) = \(\frac{1}{\sqrt{8+h-x}+\sqrt{8-x}}\)

Here, h→0

So, g’(x) = \(\frac{1}{\sqrt{8+0-x}+\sqrt{8-x}}\)

g’(x) = \(\frac{1}{\sqrt{8-x}+\sqrt{8-x}}\)

g’(x) = \(\frac{1}{2\sqrt{8-x}}\)

Therefore, the derivative of the function is g’(x) = \(\frac{1}{2\sqrt{8-x}}\).

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Find the derivative of the function using the definition of derivative. g(x) = √8 - x.

Summary:

The derivative of the function g(x) = √(8 - x) using the definition of derivative is g’(x)=\(\frac{1}{2\sqrt{8-x}}\).