मराठी

tan √ x

Advertisements
Advertisements

प्रश्न

\[\tan \sqrt{x}\]

Advertisements

उत्तर

\[Let f(x) = \tan\sqrt{x}\]
\[\text{ Thus, we have }: \]
\[(x + h) = \tan\sqrt{x + h}\]
\[\frac{d}{dx}(f(x)) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\]
\[ = \lim_{h \to 0} \frac{\tan\sqrt{x + h} - \tan\sqrt{x}}{h}\]
\[ = \lim_{h \to 0} \frac{\sin \left( \sqrt{x + h} - \sqrt{x} \right)}{h \cos\sqrt{x + h} \cos \sqrt{x}} \left[ \because \tan A - \tan B = \frac{\sin(A - B)}{\cos A \cos B} \right] \]
\[ = \lim_{h \to 0} \frac{\sin \left( \sqrt{x + h} - \sqrt{x} \right)}{\left( x + h - x \right) \cos\sqrt{x + h} \cos \sqrt{x}} \]
\[ = \lim_{h \to 0} \frac{\sin \left( \sqrt{x + h} - \sqrt{x} \right)}{\left( \sqrt{x + h} - \sqrt{x} \right)\left( \sqrt{x + h} - \sqrt{x} \right)\cos\sqrt{x + h} \cos \sqrt{x}}\]
\[ = \lim_{h \to 0} \frac{\sin \left( \sqrt{x + h} - \sqrt{x} \right)}{\left( \sqrt{x + h} - \sqrt{x} \right)} . \lim_{h \to 0} \frac{1}{\left( \sqrt{x + h} + \sqrt{x} \right)\cos\sqrt{x + h}\cos\sqrt{x}} \left[ \because \lim_{h \to 0} \frac{\sin\left( \sqrt{x + h} - \sqrt{x} \right)}{\sqrt{x + h} - \sqrt{x}} = 1 \right]\]
\[ = 1 \times \frac{1}{2\sqrt{x}\cos\sqrt{x}\cos\sqrt{x}}\]
\[ = \frac{1}{2\sqrt{x}} \sec^2 \sqrt{x}\]

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

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 30 Derivatives
Exercise 30.2 | Q 5.3 | पृष्ठ २६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the derivative of x2 – 2 at x = 10.


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


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):

`(a + bsin x)/(c + dcosx)`


Find the derivative of f (x) = x2 − 2 at x = 10


Find the derivative of f (x) = cos x at x = 0


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

 sin x at x =\[\frac{\pi}{2}\]

 


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


Differentiate  of the following from first principle:

\[\cos\left( x - \frac{\pi}{8} \right)\]


Differentiate each of the following from first principle:

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


Differentiate each of the following from first principle:

\[3^{x^2}\]


\[\tan \sqrt{x}\] 


x4 − 2 sin x + 3 cos x


3x + x3 + 33


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


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


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


Find the slope of the tangent to the curve (x) = 2x6 + x4 − 1 at x = 1.


x3 sin 


x3 ex cos 


x5 (3 − 6x−9


x4 (3 − 4x−5)


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. 

 (3x2 + 2)2


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.

(x + 2) (x + 3)

 


(ax + b) (a + d)2


\[\frac{x + e^x}{1 + \log x}\] 


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


\[\frac{x^2 - x + 1}{x^2 + x + 1}\] 


\[\frac{a + \sin x}{1 + a \sin x}\] 


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


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


\[\frac{\sec x - 1}{\sec x + 1}\] 


If x < 2, then write the value of \[\frac{d}{dx}(\sqrt{x^2 - 4x + 4)}\] 


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


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


Mark the correct alternative in each of the following: 

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


Find the derivative of 2x4 + x.


Find the derivative of x2 cosx.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×