Advertisements
Advertisements
प्रश्न
Differentiate each of the following from first principle:
\[\frac{\sin x}{x}\]
Advertisements
उत्तर
\[\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\frac{\sin \left( x + h \right)}{x + h} - \frac{\sin x}{x}}{h}\]
\[ = \lim_{h \to 0} \frac{x \sin \left( x + h \right) - \left( x + h \right) \sin x}{h x \left( x + h \right)}\]
\[ = \lim_{h \to 0} \frac{x \left( \sin x \cos h + \cos x \sin h \right) - x \sin x - h \sin x}{h x \left( x + h \right)}\]
\[ = \lim_{h \to 0} \frac{x \sin x \cos h + x \cos x \sin h - x \sin x - h \sin x}{h x \left( x + h \right)}\]
\[ = \lim_{h \to 0} \frac{x \sin x \cos h - x \sin x + x \cos x \sin h - h \sin x}{h x \left( x + h \right)}\]
\[ = x \sin x \lim_{h \to 0} \frac{\cos h - 1}{h} + \frac{x \cos x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\sin x}{x} \lim_{h \to 0} \frac{1}{x + h}\]
\[ = x \sin x \lim_{h \to 0} \frac{- 2 \sin^2 \frac{h}{2}}{h} + \frac{x \cos x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\sin x}{x} \lim_{h \to 0} \frac{1}{x + h}\]
\[ = x \sin x \lim_{h \to 0} \frac{- 2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4} + \frac{x \cos x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\sin x}{x} \lim_{h \to 0} \frac{1}{x + h}\]
\[ = - x \sin x \times \lim_{h \to 0} \frac{h}{2} + \frac{x \cos x}{x} \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{x + h} - \frac{\sin x}{x} \lim_{h \to 0} \frac{1}{x + h}\]
\[ = - x \sin x \left( \frac{1}{2} \right) \left( 0 \right) + \frac{cos x}{x} - \frac{sin x}{x^2}\]
\[ = \frac{\cos x}{x} - \frac{\sin x}{x^2}\]
\[ = \frac{x \cos x - \sin x}{x^2}\]
APPEARS IN
संबंधित प्रश्न
Find the derivative of x2 – 2 at x = 10.
Find the derivative of x at x = 1.
For the function
f(x) = `x^100/100 + x^99/99 + ...+ x^2/2 + x + 1`
Prove that f'(1) = 100 f'(0)
Find the derivative of the following function (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):
`1/(ax^2 + bx + c)`
Find the derivative of the following function (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):
`(sin x + cos x)/(sin x - cos x)`
Find the derivative of f (x) = cos x at x = 0
\[\frac{1}{\sqrt{3 - x}}\]
(x2 + 1) (x − 5)
Differentiate of the following from first principle:
e3x
Differentiate of the following from first principle:
\[\cos\left( x - \frac{\pi}{8} \right)\]
tan2 x
tan 2x
\[\sqrt{\tan x}\]
\[\tan \sqrt{x}\]
\[\frac{x^3}{3} - 2\sqrt{x} + \frac{5}{x^2}\]
(2x2 + 1) (3x + 2)
\[\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^3\]
\[\frac{a \cos x + b \sin x + c}{\sin x}\]
a0 xn + a1 xn−1 + a2 xn−2 + ... + an−1 x + an.
\[\log\left( \frac{1}{\sqrt{x}} \right) + 5 x^a - 3 a^x + \sqrt[3]{x^2} + 6 \sqrt[4]{x^{- 3}}\]
\[If y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}, \text{ prove that } 2xy\frac{dy}{dx} = \left( \frac{x}{a} - \frac{a}{x} \right)\]
x3 sin x
x3 ex
\[\frac{2^x \cot x}{\sqrt{x}}\]
x−3 (5 + 3x)
\[\frac{x^2 + 1}{x + 1}\]
\[\frac{x}{1 + \tan x}\]
\[\frac{1}{a x^2 + bx + c}\]
\[\frac{x^2 - x + 1}{x^2 + x + 1}\]
\[\frac{3^x}{x + \tan x}\]
Write the value of \[\frac{d}{dx}\left( x \left| x \right| \right)\]
Mark the correct alternative in of the following:
If\[f\left( x \right) = 1 - x + x^2 - x^3 + . . . - x^{99} + x^{100}\]then \[f'\left( 1 \right)\]
Mark the correct alternative in each of the following:
If\[y = \frac{\sin x + \cos x}{\sin x - \cos x}\] then \[\frac{dy}{dx}\]at x = 0 is
Mark the correct alternative in of the following:
If \[y = \frac{\sin\left( x + 9 \right)}{\cos x}\] then \[\frac{dy}{dx}\] at x = 0 is
Mark the correct alternative in of the following:
If f(x) = x sinx, then \[f'\left( \frac{\pi}{2} \right) =\]
Find the derivative of f(x) = tan(ax + b), by first principle.
(ax2 + cot x)(p + q cos x)
Let f(x) = x – [x]; ∈ R, then f'`(1/2)` is ______.
