English

Tan 2x

Advertisements
Advertisements

Question

 tan 2

Advertisements

Solution

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 30: Derivatives - Exercise 30.2 [Page 26]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 30 Derivatives
Exercise 30.2 | Q 4.3 | Page 26

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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):

`4sqrtx - 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):

(ax + b)n (cx + d)m


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


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


Differentiate  of the following from first principle:

 x sin x


Differentiate of the following from first principle:

 x cos x


Differentiate each of the following from first principle:

x2 e


Differentiate each of the following from first principle:

\[3^{x^2}\]


\[\cos \sqrt{x}\]


(2x2 + 1) (3x + 2) 


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


a0 xn + a1 xn−1 + a2 xn2 + ... + an1 x + an


\[If y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}, \text{ prove that } 2xy\frac{dy}{dx} = \left( \frac{x}{a} - \frac{a}{x} \right)\]  


x3 e


\[\frac{2^x \cot x}{\sqrt{x}}\] 


x5 ex + x6 log 


(1 +x2) cos x


\[e^x \log \sqrt{x} \tan x\] 


Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.

(3 sec x − 4 cosec x) (−2 sin x + 5 cos x)


(ax + b)n (cx d)


\[\frac{x^2 + 1}{x + 1}\] 


\[\frac{2x - 1}{x^2 + 1}\] 


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


\[\frac{a x^2 + bx + c}{p x^2 + qx + r}\] 


\[\frac{e^x + \sin x}{1 + \log x}\] 


\[\frac{{10}^x}{\sin x}\] 


\[\frac{4x + 5 \sin x}{3x + 7 \cos x}\]


\[\frac{x}{\sin^n x}\]


\[\frac{1}{a x^2 + bx + c}\] 


Write the value of \[\frac{d}{dx} \left\{ \left( x + \left| x \right| \right) \left| x \right| \right\}\]


If f (x) = |x| + |x−1|, write the value of \[\frac{d}{dx}\left( f (x) \right)\]


If |x| < 1 and y = 1 + x + x2 + x3 + ..., then write the value of \[\frac{dy}{dx}\] 


Mark the correct alternative in of the following: 

If \[f\left( x \right) = \frac{x - 4}{2\sqrt{x}}\]

 


Mark the correct alternative in of the following:
If \[f\left( x \right) = x^{100} + x^{99} + . . . + x + 1\]  then \[f'\left( 1 \right)\] is equal to 


Mark the correct alternative in each of the following: 

If\[f\left( x \right) = \frac{x^n - a^n}{x - a}\] then \[f'\left( a \right)\] 


`(a + b sin x)/(c + d cos x)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×