Advertisements
Advertisements
Question
Differentiate each of the following from first principle:
\[\sqrt{\sin (3x + 1)}\]
Advertisements
Solution
\[ \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\sqrt{\sin \left( 3\left( x + h \right) + 1 \right)} - \sqrt{\sin \left( 3x + 1 \right)}}{h} \]
\[ = \lim_{h \to 0} \frac{\sqrt{\sin \left( 3x + 3h + 1 \right)} - \sqrt{\sin \left( 3x + 1 \right)}}{h} \times \frac{\sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)}}{\sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)}}\]
\[ = \lim_{h \to 0} \frac{\sin \left( 3x + 3h + 1 \right) - \sin \left( 3x + 1 \right)}{h \left( \sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)} \right)}\]
\[We have:\]
\[ sin C-sin D= 2 cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right)\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{3x + 3h + 1 + 3x + 1}{2} \right) \sin \left( \frac{3x + 3h + 1 - 3x - 1}{2} \right)}{h \left( \sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)} \right)}\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{6x + 3h + 2}{2} \right) \sin \frac{3h}{2}}{h \left( \sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)} \right)}\]
\[ = \lim_{h \to 0} 2 \cos \left( \frac{6x + 3h + 2}{2} \right) \lim_{h \to 0} \frac{\sin \frac{3h}{2}}{h \times \frac{3}{2}} \times \frac{3}{2} \times \lim_{h \to 0} \frac{1}{\left( \sqrt{\sin \left( 3x + 3h + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)} \right)} \]
\[ = 2 \cos \left( 3x + 1 \right) \times \left( \frac{3}{2} \right) \times \frac{1}{\sqrt{\sin \left( 3x + 1 \right)} + \sqrt{\sin \left( 3x + 1 \right)}}\]
\[ = \frac{3 \cos \left( 3x + 1 \right)}{2\sqrt{\sin \left( 3x + 1 \right)}}\]
\[ \]
\[\]
APPEARS IN
RELATED QUESTIONS
Find the derivative of x2 – 2 at x = 10.
Find the derivative of 99x at x = 100.
Find the derivative of x–4 (3 – 4x–5).
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):
`4sqrtx - 2`
\[\frac{x^2 - 1}{x}\]
\[\frac{1}{\sqrt{3 - x}}\]
(x2 + 1) (x − 5)
\[\sqrt{2 x^2 + 1}\]
Differentiate of the following from first principle:
sin (x + 1)
Differentiate of the following from first principle:
sin (2x − 3)
Differentiate each of the following from first principle:
\[\frac{\sin x}{x}\]
Differentiate each of the following from first principle:
\[e^\sqrt{2x}\]
Differentiate each of the following from first principle:
\[3^{x^2}\]
tan 2x
\[\tan \sqrt{x}\]
x4 − 2 sin x + 3 cos x
ex log a + ea long x + ea log a
log3 x + 3 loge x + 2 tan x
\[\left( x + \frac{1}{x} \right)\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\]
\[\frac{( x^3 + 1)(x - 2)}{x^2}\]
\[\frac{a \cos x + b \sin x + c}{\sin x}\]
\[\text{ If } y = \left( \frac{2 - 3 \cos x}{\sin x} \right), \text{ find } \frac{dy}{dx} at x = \frac{\pi}{4}\]
x3 ex
xn loga x
x2 sin x log x
x5 ex + x6 log x
\[e^x \log \sqrt{x} \tan x\]
(ax + b) (a + d)2
(ax + b)n (cx + d)n
\[\frac{a x^2 + bx + c}{p x^2 + qx + r}\]
\[\frac{x \tan x}{\sec x + \tan x}\]
\[\frac{x \sin x}{1 + \cos x}\]
\[\frac{a + \sin x}{1 + a \sin x}\]
\[\frac{x}{1 + \tan x}\]
\[\frac{\sec x - 1}{\sec x + 1}\]
Write the value of \[\frac{d}{dx}\left( x \left| x \right| \right)\]
If f (x) = \[\frac{x^2}{\left| x \right|},\text{ write }\frac{d}{dx}\left( f (x) \right)\]
Mark the correct alternative in of the following:
If \[y = \frac{1 + \frac{1}{x^2}}{1 - \frac{1}{x^2}}\] then \[\frac{dy}{dx} =\]
Mark the correct alternative in of the following:
If \[y = \sqrt{x} + \frac{1}{\sqrt{x}}\] then \[\frac{dy}{dx}\] at x = 1 is
