मराठी

Tan 2x - Mathematics

Advertisements
Advertisements

प्रश्न

 tan 2

Advertisements

उत्तर

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 30: Derivatives - Exercise 30.2 [पृष्ठ २६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 30 Derivatives
Exercise 30.2 | Q 4.3 | पृष्ठ २६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the derivative of 99x at x = 100.


Find the derivative of (5x3 + 3x – 1) (x – 1).


Find the derivative of x5 (3 – 6x–9).


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 at the indicated point:

2 cos x at x =\[\frac{\pi}{2}\] 


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

 sin 2x at x =\[\frac{\pi}{2}\]


k xn


(x + 2)3


Differentiate  of the following from first principle:

 eax + b


Differentiate of the following from first principle:

 x cos x


Differentiate each of the following from first principle:

\[\sqrt{\sin 2x}\] 


Differentiate each of the following from first principle: 

sin x + cos x


Differentiate each of the following from first principle: 

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


Differentiate each  of the following from first principle:

\[e^\sqrt{2x}\]


tan2 


\[\sqrt{\tan x}\]


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


(2x2 + 1) (3x + 2) 


 log3 x + 3 loge x + 2 tan x


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


cos (x + a)


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


(x3 + x2 + 1) sin 


\[\frac{2^x \cot x}{\sqrt{x}}\] 


x2 sin x log 


(1 +x2) cos x


\[\frac{x + e^x}{1 + \log x}\] 


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


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


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


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


\[\frac{a + \sin x}{1 + a \sin x}\] 


\[\frac{1 + \log x}{1 - \log x}\] 


\[\frac{p x^2 + qx + r}{ax + b}\]


\[\frac{x + \cos x}{\tan x}\] 


\[\frac{ax + b}{p x^2 + qx + r}\] 


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


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


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


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 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×