Find the derivative of \(\sqrt{4-x}\) from first principle.

1.	Find the derivative of \(\sqrt{4-x}\) from first principle.
Answer» Let, \(f(x)=\sqrt{4-x}\) ∴ \(f'(x)=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\) \(=\lim\limits_{h \to 0}\frac{\sqrt{4-(x+h)}-\sqrt{4-x}}{h}\) \(=\lim\limits_{h \to 0}\frac{[\sqrt{4-(x+h)}-\sqrt{4-x}][\sqrt{4-(x+h)}+\sqrt{4-x}]}{[h\sqrt{4-(x+h)}+\sqrt{4-x}]}\) \(=\lim\limits_{h \to 0}\frac{[{4-(x+h)}]-(4-x)}{h[\sqrt{4-(x+h)}+\sqrt{4-x}]}\) \(=\lim\limits_{h \to 0}\frac{-h}{h\sqrt{4-(x+h)}+\sqrt{4-x}}\) \(=\lim\limits_{h \to 0}\frac{1}{\sqrt{4-(x+h)}+\sqrt{4-x}}\) = \(-\frac{1}{2\sqrt{4-x}}\)

Answer»

Let,

\(f(x)=\sqrt{4-x}\)

∴ \(f'(x)=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\)

\(=\lim\limits_{h \to 0}\frac{\sqrt{4-(x+h)}-\sqrt{4-x}}{h}\)

\(=\lim\limits_{h \to 0}\frac{[\sqrt{4-(x+h)}-\sqrt{4-x}][\sqrt{4-(x+h)}+\sqrt{4-x}]}{[h\sqrt{4-(x+h)}+\sqrt{4-x}]}\)

\(=\lim\limits_{h \to 0}\frac{[{4-(x+h)}]-(4-x)}{h[\sqrt{4-(x+h)}+\sqrt{4-x}]}\)

\(=\lim\limits_{h \to 0}\frac{-h}{h\sqrt{4-(x+h)}+\sqrt{4-x}}\)

\(=\lim\limits_{h \to 0}\frac{1}{\sqrt{4-(x+h)}+\sqrt{4-x}}\)

= \(-\frac{1}{2\sqrt{4-x}}\)

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