हिंदी

Differentiate X X Cos X + X 2 + 1 X 2 − 1 ? - Mathematics

Advertisements
Advertisements

प्रश्न

Differentiate  \[x^{x \cos x +} \frac{x^2 + 1}{x^2 - 1}\]  ?

Advertisements

उत्तर

\[\text{ Let y }= x^{x \cos x} + \frac{x^2 + 1}{x^2 - 1}\]

\[\text{ Also, Let u } = x^{x \cos x} \text{ and v } = \frac{x^2 + 1}{x^2 - 1}\]

\[ \therefore y = u + v\]

\[ \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} . . . \left( i \right)\]

\[\text{ Now, u }= x^{x \cos x} \]

\[ \Rightarrow \log u = \log\left( x^{x \cos x} \right)\]

\[ \Rightarrow \log u = x \cos x \log x\]

Differentiating both sides with respect to x,

\[\frac{1}{u}\frac{du}{dx} = \cos x \log x\frac{d}{dx}\left( x \right) + x\log x\frac{d}{dx}\left( \cos x \right) + x \cos x\frac{d}{dx}\left( \log x \right)\]

\[ \Rightarrow \frac{du}{dx} = u\left[ \cos x \log x + x\left( - \sin x \right)\log x + x \cos x\left( \frac{1}{x} \right) \right]\]

\[ \Rightarrow \frac{du}{dx} = x^{x \cos x} \left( \cos x \log x - x \sin x \log x + \cos x \right)\]

\[ \Rightarrow \frac{du}{dx} = x^{x \cos x} \left[ \cos x\left( 1 + \log x \right) - x \sin x \log x \right] . . . \left( 2 \right)\]

\[\text{ Again, v }= \frac{x^2 + 1}{x^2 - 1}\]

\[ \Rightarrow \log v = \log\left( x^2 + 1 \right) - \log\left( x^2 - 1 \right)\]

Differentiating both sides with respect to x,

\[\frac{1}{v}\frac{dv}{dx} = \frac{2x}{x^2 + 1} - \frac{2x}{x^2 - 1}\]

\[ \Rightarrow \frac{dv}{dx} = v\left[ \frac{2x\left( x^2 - 1 \right) - 2x\left( x^2 + 1 \right)}{\left( x^2 + 1 \right)\left( x^2 - 1 \right)} \right]\]

\[ \Rightarrow \frac{dv}{dx} = \frac{x^2 + 1}{x^2 - 1}\left[ \frac{- 4x}{\left( x^2 + 1 \right)\left( x^2 - 1 \right)} \right]\]

\[ \Rightarrow \frac{dv}{dx} = \frac{- 4x}{\left( x^2 - 1 \right)^2} . . . \left( 3 \right)\]

\[\text{ From} \left( i \right), \left( ii \right) \text{ and } \left( iii \right), \text{ we obtain}\]

\[\frac{dy}{dx} = x^{x \cos x} \left[ \cos x\left( 1 + \log x \right) - x \sin x \log x \right] - \frac{4x}{\left( x^2 - 1 \right)^2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Differentiation - Exercise 11.05 [पृष्ठ ८८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.05 | Q 18.3 | पृष्ठ ८८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Differentiate the following functions from first principles ecos x.


Differentiate sin2 (2x + 1) ?


Differentiate \[\log \left( \frac{\sin x}{1 + \cos x} \right)\] ?


Differentiate \[\log \left( cosec x - \cot x \right)\] ?


Differentiate \[\frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}}\] ?


If  \[y = \left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)\] , prove that \[\frac{dy}{dc} = \log \left( \frac{x - 1}{1 + x} \right)\] ?


Differentiate \[\tan^{- 1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\}, - 1 < x < 1\] ?


Differentiate \[\sin^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?


Find  \[\frac{dy}{dx}\] in the following case: \[y^3 - 3x y^2 = x^3 + 3 x^2 y\] ?

 


If \[\log \sqrt{x^2 + y^2} = \tan^{- 1} \left( \frac{y}{x} \right)\] Prove that \[\frac{dy}{dx} = \frac{x + y}{x - y}\] ?


If \[\sec \left( \frac{x + y}{x - y} \right) = a\] Prove that  \[\frac{dy}{dx} = \frac{y}{x}\] ?


If  \[\tan \left( x + y \right) + \tan \left( x - y \right) = 1, \text{ find}  \frac{dy}{dx}\] ?


If \[\sin^2 y + \cos xy = k,\] find  \[\frac{dy}{dx}\] at \[x = 1 , \] \[y = \frac{\pi}{4} .\] 


If \[\sqrt{y + x} + \sqrt{y - x} = c, \text {show that } \frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}\] ?


Find  \[\frac{dy}{dx}\]  \[y = \frac{e^{ax} \cdot \sec x \cdot \log x}{\sqrt{1 - 2x}}\] ?

 


Find  \[\frac{dy}{dx}\] \[y = \sin x \sin 2x \sin 3x \sin 4x\] ?

 


Find the derivative of the function f (x) given by  \[f\left( x \right) = \left( 1 + x \right) \left( 1 + x^2 \right) \left( 1 + x^4 \right) \left( 1 + x^8 \right)\] and hence find `f' (1)` ?

 


If \[y = \left( \sin x - \cos x \right)^{\sin x - \cos x} , \frac{\pi}{4} < x < \frac{3\pi}{4}, \text{ find} \frac{dy}{dx}\] ?


\[\text{ If } x = e^{x/y} , \text{ prove that } \frac{dy}{dx} = \frac{x - y}{x\log x}\] ?

If \[y = \left( \tan x \right)^{\left( \tan x \right)^{\left( \tan x \right)^{. . . \infty}}}\], prove that \[\frac{dy}{dx} = 2\ at\ x = \frac{\pi}{4}\] ?

 


Find \[\frac{dy}{dx}\], when \[x = a t^2 \text{ and } y = 2\ at \] ?


Find \[\frac{dy}{dx}\] ,When \[x = a \left( 1 - \cos \theta \right) \text{ and } y = a \left( \theta + \sin \theta \right) \text{ at } \theta  = \frac{\pi}{2}\] ?


If  \[x = a\left( t + \frac{1}{t} \right) \text{ and y } = a\left( t - \frac{1}{t} \right)\] ,prove that  \[\frac{dy}{dx} = \frac{x}{y}\]?

 


If  \[x = \frac{\sin^3 t}{\sqrt{\cos 2 t}}, y = \frac{\cos^3 t}{\sqrt{\cos t 2 t}}\] , find\[\frac{dy}{dx}\] ?

 


If \[y = \sin^{- 1} \left( \sin x \right), - \frac{\pi}{2} \leq x \leq \frac{\pi}{2}\] ,Then, write the value of \[\frac{dy}{dx} \text{ for } x \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \] ?


If \[x = a \left( \theta + \sin \theta \right), y = a \left( 1 + \cos \theta \right), \text{ find} \frac{dy}{dx}\] ?


If \[- \frac{\pi}{2} < x < 0 \text{ and y } = \tan^{- 1} \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}, \text{ find } \frac{dy}{dx}\] ?


Let  \[\cup = \sin^{- 1} \left( \frac{2 x}{1 + x^2} \right) \text { and }V = \tan^{- 1} \left( \frac{2 x}{1 - x^2} \right), \text { then } \frac{d \cup}{dV} =\] ____________ .


If \[y = \log \left( \frac{1 - x^2}{1 + x^2} \right), \text { then } \frac{dy}{dx} =\] __________ .


Find the second order derivatives of the following function tan−1 x ?


If y = log (sin x), prove that \[\frac{d^3 y}{d x^3} = 2 \cos \ x \ {cosec}^3 x\] ?


If x = cos θ, y = sin3 θ, prove that \[y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 = 3 \sin^2 \theta\left( 5 \cos^2 \theta - 1 \right)\] ?


If x = sin ty = sin pt, prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] ?


Find \[\frac{d^2 y}{d x^2}\] where \[y = \log \left( \frac{x^2}{e^2} \right)\] ?


If y = ex (sin + cos x) prove that \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\] ?


\[\text { If y } = x^n \left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\}, \text { prove that } x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)x\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0 \] Disclaimer: There is a misprint in the question. It must be 

\[x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)x\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0\] instead of 1

\[x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0\] ?


If \[f\left( x \right) = \frac{\sin^{- 1} x}{\sqrt{1 - x^2}}\] then (1 − x)2 '' (x) − xf(x) =

 


If y = etan x, then (cos2 x)y2 =


If y = xn−1 log x then x2 y2 + (3 − 2n) xy1 is equal to


If y2 = ax2 + bx + c, then \[y^3 \frac{d^2 y}{d x^2}\] is 

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×