हिंदी

Differentiate of the Following from First Principle: X Cos X

Advertisements
Advertisements

प्रश्न

Differentiate of the following from first principle:

 x cos x

Advertisements

उत्तर

\[\left( x \right) \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\left( x + h \right) \cos \left( x + h \right) - x \cos x}{h}\]
\[ = \lim_{h \to 0} \frac{\left( x + h \right)\left( \cos x \cos h - \sin x \sin h \right) - x \cos x}{h}\]
\[ = \lim_{h \to 0} \frac{x \cos x \cos h - x \sin x \sin h + h \cos x \cos h - h \sin x \sin h - x \cos x}{h}\]
\[ = \lim_{h \to 0} \frac{x \cos x \cos h - x \cos x - x \sin x \sin h + h \cos x \cos h - h \sin x \sin h}{h}\]
\[ = x \cos x \lim_{h \to 0} \frac{\left( \cos h - 1 \right)}{h} - x \sin x \lim_{h \to 0} \frac{\sin h}{h} + \cos x \lim_{h \to 0} \cos h + \sin x \lim_{h \to 0} \sin h\]
\[ = x \cos x \lim_{h \to 0} \frac{- 2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4} - x \sin x \left( 1 \right) + \cos x \left( 1 \right) + \sin x \left( 0 \right)\]
\[ = x\cos x \lim_{h \to 0} \frac{- h}{2} - x \sin x \left( 1 \right) + \cos x \left( 1 \right) + \sin x \left( 0 \right)\]
\[ = - x \cos x \left( 0 \right) - x \sin x + \cos x \]
\[ = - x \sin x + \cos x \]
\[ \]
\[\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 30: Derivatives - Exercise 30.2 [पृष्ठ २५]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 30 Derivatives
Exercise 30.2 | Q 2.1 | पृष्ठ २५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find the derivative of x at x = 1.


Find the derivative of `2x - 3/4`


Find the derivative of `2/(x + 1) - x^2/(3x -1)`.


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):

(x + a)


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):

`(ax + b)/(px^2 + qx + r)`


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) = 3x at x = 2 


Find the derivative of the following function at the indicated point:


Find the derivative of the following function at the indicated point: 

 sin 2x at x =\[\frac{\pi}{2}\]


\[\frac{1}{\sqrt{x}}\]


\[\sqrt{2 x^2 + 1}\]


tan2 


x4 − 2 sin x + 3 cos x


3x + x3 + 33


\[\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^3\] 


\[\frac{a \cos x + b \sin x + c}{\sin x}\]


2 sec x + 3 cot x − 4 tan x


\[\text{ If } y = \left( \sin\frac{x}{2} + \cos\frac{x}{2} \right)^2 , \text{ find } \frac{dy}{dx} at x = \frac{\pi}{6} .\]


\[\text{ If } y = \frac{2 x^9}{3} - \frac{5}{7} x^7 + 6 x^3 - x, \text{ find } \frac{dy}{dx} at x = 1 .\] 


xn loga 


(x sin x + cos x ) (ex + x2 log x


(2x2 − 3) sin 


x4 (3 − 4x−5)


\[\frac{x + e^x}{1 + \log x}\] 


\[\frac{e^x - \tan x}{\cot x - x^n}\] 


\[\frac{e^x}{1 + x^2}\] 


\[\frac{x \tan x}{\sec x + \tan x}\]


\[\frac{3^x}{x + \tan x}\] 


\[\frac{1 + \log x}{1 - \log x}\] 


\[\frac{x + \cos x}{\tan x}\] 


Write the value of the derivative of f (x) = |x − 1| + |x − 3| at x = 2.


If f (x) = \[\frac{x^2}{\left| x \right|},\text{ write }\frac{d}{dx}\left( f (x) \right)\] 


If f (1) = 1, f' (1) = 2, then write the value of \[\lim_{x \to 1} \frac{\sqrt{f (x)} - 1}{\sqrt{x} - 1}\] 


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  of the following:
If\[f\left( x \right) = 1 + x + \frac{x^2}{2} + . . . + \frac{x^{100}}{100}\] then \[f'\left( 1 \right)\] is equal to 


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 


Let f(x) = x – [x]; ∈ R, then f'`(1/2)` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×