Differentiate-sin-x-from-the-first-principle-

Question Number 9231 by tawakalitu last updated on 24/Nov/16

Differentiate : \sin \sqrt{x} from the

first principle .

Answered by mrW last updated on 29/Nov/16

y (x) = \sin \sqrt{x}

y (x + h) = \sin \sqrt{x + h}

\frac{y (x + h) - y (x)}{h} = \frac{\sin \sqrt{x + h} - \sin \sqrt{x}}{h}

= \frac{2 \times \cos \frac{\sqrt{x + h} + \sqrt{x}}{2} \times \sin \frac{\sqrt{x + h} - \sqrt{x}}{2}}{h}

= 2 \times \cos \frac{\sqrt{x + h} + \sqrt{x}}{2} \times \frac{\sin \frac{\sqrt{x + h} - \sqrt{x}}{2}}{h}

let u = \frac{\sqrt{x + h} - \sqrt{x}}{2}

\sqrt{x + h} - \sqrt{x} = 2 u

\sqrt{x + h} = 2 u + \sqrt{x}

h = {(2 u + \sqrt{x})}^{2} - {(\sqrt{x})}^{2} = 4 u (u + \sqrt{x})

\frac{\sin \frac{\sqrt{x + h} - \sqrt{x}}{2}}{h} = \frac{\sin u}{4 u (u + \sqrt{x)}} = \frac{1}{4} \times \frac{\sin u}{u} \times \frac{1}{u + \sqrt{x}}

with h \to 0, u \to 0

\lim_{h \to 0} \frac{\sin \frac{\sqrt{x + h} - \sqrt{x}}{2}}{h} = \frac{1}{4} \times \lim_{u \to 0} \frac{\sin u}{u} \times \lim_{u \to 0} \frac{1}{u + \sqrt{x}} = \frac{1}{4} \times 1 \times \frac{1}{\sqrt{x}} = \frac{1}{4 \sqrt{x}}

\frac{dy}{dx} = \lim_{h \to 0} \frac{y (x + h) - y (x)}{h}

= \lim_{h \to 0} [2 \times \cos \frac{\sqrt{x + h} + \sqrt{x}}{2} \times \frac{\sin \frac{\sqrt{x + h} - \sqrt{x}}{2}}{h}]

= 2 \times \underset{h \to 0}{\lim c o s} \frac{\sqrt{x + h} + \sqrt{x}}{2} \times \lim_{h \to 0} \frac{\sin \frac{\sqrt{x + h} - \sqrt{x}}{2}}{h}

= 2 \times \cos \sqrt{x} \times \frac{1}{4 \sqrt{x}} = \frac{\cos \sqrt{x}}{2 \sqrt{x}}

Commented by tawakalitu last updated on 29/Nov/16

Thanks so much . God bless you .