English

√ 2 X 2 + 1 - Mathematics

Advertisements
Advertisements

Question

 (x2 + 1) (x − 5)

Solution

\[ \left( x^2 + 1 \right)\left( x - 5 \right) = x^3 - 5 x^2 + x - 5\]
\[\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\left( x + h \right)^3 - 5 \left( x + h \right)^2 + \left( x + h \right) - 5 - \left( x^3 - 5 x^2 + x - 5 \right)}{h}\]
\[ = \lim_{h \to 0} \frac{x^3 + 3 x^2 h + 3x h^2 + h^3 - 5 x^2 - 5 h^2 - 10xh + x + h - 5 - x^3 + 5 x^2 - x + 5}{h}\]
\[ = \lim_{h \to 0} \frac{3 x^2 h + 3x h^2 + h^3 - 5 h^2 - 10xh + h}{h}\]
\[ = \lim_{h \to 0} \frac{h \left( 3 x^2 + 3xh + h^2 - 5h - 10x + 1 \right)}{h}\]
\[ = \lim_{h \to 0} \left( 3 x^2 + 3xh + h^2 - 5h - 10x + 1 \right)\]
\[ = 3 x^2 - 10x + 1\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 30: Derivatives - Exercise 30.2 [Page 25]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 30 Derivatives
Exercise 30.2 | Q 1.13 | Page 25

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find the derivative of 99x at x = 100.


Find the derivative of x at x = 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) (cx + d)2


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):

x4 (5 sin x – 3 cos x)


Find the derivative of f (x) = 3x at x = 2 


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


\[\frac{2}{x}\]


\[\frac{1}{x^3}\]


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


Differentiate of the following from first principle:

(−x)−1


Differentiate  of the following from first principle:

sin (x + 1)


Differentiate each of the following from first principle:

\[\sqrt{\sin (3x + 1)}\]


\[\sqrt{\tan x}\]


\[\frac{x^3}{3} - 2\sqrt{x} + \frac{5}{x^2}\]


\[\frac{2 x^2 + 3x + 4}{x}\] 


\[\frac{( x^3 + 1)(x - 2)}{x^2}\] 


\[\frac{(x + 5)(2 x^2 - 1)}{x}\]


cos (x + a)


(x sin x + cos x ) (ex + x2 log x


\[e^x \log \sqrt{x} \tan x\] 


x3 ex cos 


(2x2 − 3) sin 


x−3 (5 + 3x


\[\frac{x + e^x}{1 + \log x}\] 


\[\frac{x}{1 + \tan x}\] 


\[\frac{1}{a x^2 + bx + c}\] 


\[\frac{x \sin x}{1 + \cos x}\]


\[\frac{\sqrt{a} + \sqrt{x}}{\sqrt{a} - \sqrt{x}}\] 


\[\frac{a + \sin x}{1 + a \sin x}\] 


\[\frac{1}{a x^2 + bx + c}\] 


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\[y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + . . .\]then \[\frac{dy}{dx} =\] 

 


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×