हिंदी

4 X + 5 Sin X 3 X + 7 Cos X

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\text{ Let } u = 4x + 5 \sin x; v = 3x + 7 \cos x\]
\[\text{ Then }, u' = 4 + 5 \cos x; v' = 3 - 7 \sin x\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{4x + 5 \sin x}{3x + 7 \cos x} \right) = \frac{\left( 3x + 7 \cos x \right)\left( 4 + 5 \cos x \right) - \left( 4x + 5 \sin x \right)\left( 3 - 7 \sin x \right)}{\left( 3x + 7 \cos x \right)^2}\]
\[ = \frac{12x + 15 x \cos x + 28 \cos x + 35 \cos^2 x - 12x + 28 x \sin x - 15 \sin x + 35 \sin^2 x}{\left( 3x + 7 \cos x \right)^2}\]
\[ = \frac{15 x \cos x + 28 x \sin x + 28 \cos x15 \sin x + 35\left( \sin^2 x + \cos^2 x \right)}{\left( 3x + 7 \cos x \right)^2}\]
\[ = \frac{15 x \cos x + 28 x \sin x + 28 \cos x15 \sin x + 35}{\left( 3x + 7 \cos x \right)^2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 30: Derivatives - Exercise 30.5 [पृष्ठ ४४]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 30 Derivatives
Exercise 30.5 | Q 21 | पृष्ठ ४४

वीडियो 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):

(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):

`(px+ q) (r/s + s)`


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)`


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):

`cos x/(1 + sin x)`


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 the following function at the indicated point: 

 sin 2x at x =\[\frac{\pi}{2}\]


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


\[\frac{2x + 3}{x - 2}\] 


Differentiate  of the following from first principle:

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


tan2 


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


\[\frac{2 x^2 + 3x + 4}{x}\] 


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


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


x2 ex log 


x2 sin x log 


logx2 x


\[\frac{x^2 \cos\frac{\pi}{4}}{\sin x}\] 


x4 (5 sin x − 3 cos x)


\[\frac{2x - 1}{x^2 + 1}\] 


\[\frac{x + e^x}{1 + \log x}\] 


\[\frac{e^x}{1 + x^2}\] 


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


\[\frac{\sec x - 1}{\sec x + 1}\] 


Write the value of \[\lim_{x \to c} \frac{f(x) - f(c)}{x - c}\] 


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


If \[\frac{\pi}{2}\] then find \[\frac{d}{dx}\left( \sqrt{\frac{1 + \cos 2x}{2}} \right)\]


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


Mark the correct alternative in of the following:

Let f(x) = x − [x], x ∈ R, then \[f'\left( \frac{1}{2} \right)\]


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 \[f'\left( a \right)\] 


Find the derivative of 2x4 + x.


Find the derivative of x2 cosx.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×