मराठी

Differentiate ( Cos X ) Sin X with Respect to ( Sin X ) Cos X ? - Mathematics

Advertisements
Advertisements

प्रश्न

Differentiate \[\left( \cos x \right)^{\sin x }\] with respect to \[\left( \sin x \right)^{\cos x }\]?

बेरीज
Advertisements

उत्तर

\[\text { Let, u } = \left( \cos x \right)^{\sin x} \]

Taking log on both sides,

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

\[ \Rightarrow \log u = \sin x \log\left( \cos x \right)\]

Differentiating it with respect to x using chain rule,

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

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

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

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

\[\text { Let, v }= \left( \sin x \right)^{\cos x }\]

Taking log on both sides,

\[\log v = \log \left( \sin x \right)^{\cos x} \]

\[ \Rightarrow \log v = \cos x \log\left( \sin x \right)\]

Differentiating it with respect to x using chain rule,

\[\frac{1}{v}\frac{dv}{dx} = \cos x\frac{d}{dx}\left( \log\sin x \right) + \log\sin x\frac{d}{dx}\left( \cos x \right) ..........\left[ \text { using product rule } \right]\]

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

\[ \Rightarrow \frac{dv}{dx} = v\left[ \cot x\left( \cos x \right) - \sin x \log\sin x \right]\]

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

\[\text { dividing equation }\left( i \right) by \left( ii \right), \]

\[ \therefore \frac{du}{dv} = \frac{\left( \cos x \right)^{\sin x } \left[ \cos x \log\cos x - \sin x \tan x \right]}{\left( \sin x \right)^{\cos x } \left[ \cot x\left( \cos x \right) - \sin x \log\sin x \right]}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Differentiation - Exercise 11.08 [पृष्ठ ११२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.08 | Q 8 | पृष्ठ ११२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Differentiate the following functions from first principles  \[e^\sqrt{2x}\].


Differentiate \[e^{\sin^{- 1} 2x}\] ?


Differentiate \[\frac{e^x \sin x}{\left( x^2 + 2 \right)^3}\] ?


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


Differentiate \[\sin^{- 1} \left\{ \frac{x}{\sqrt{x^2 + a^2}} \right\}\] ?


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


Differentiate \[\tan^{- 1} \left( \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \right), x \neq 0\] ?


Differentiate \[\tan^{- 1} \left( \frac{x}{1 + 6 x^2} \right)\] ?


Differentiate the following with respect to x

\[\cos^{- 1} \left( \sin x \right)\]


If  \[y = \cot^{- 1} \left\{ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right\}\],  show that \[\frac{dy}{dx}\] is independent of x. ? 

 


If  \[y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}, 0 < x < \frac{1}{2}, \text{ find } \frac{dy}{dx} .\] ?


Find  \[\frac{dy}{dx}\] in the following case \[e^{x - y} = \log \left( \frac{x}{y} \right)\] ?

 


If \[y = \left\{ \log_{\cos x} \sin x \right\} \left\{ \log_{\sin x} \cos x \right\}^{- 1} + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right), \text{ find } \frac{dy}{dx} \text{ at }x = \frac{\pi}{4}\] ?


Differentiate \[\left( \log x \right)^{\cos x}\] ?


Differentiate  \[\sin \left( x^x \right)\] ?


Differentiate \[\left( \tan x \right)^{1/x}\] ?


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


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


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

 


Find \[\frac{dy}{dx}\] \[y = x^{\log x }+ \left( \log x \right)^x\] ?


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


If \[x^m y^n = 1\] , prove that \[\frac{dy}{dx} = - \frac{my}{nx}\] ?


Find \[\frac{dy}{dx}\],when \[x = a e^\theta \left( \sin \theta - \cos \theta \right), y = a e^\theta \left( \sin \theta + \cos \theta \right)\] ?


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 \[x = \cos t \text{ and y }  = \sin t,\] prove that  \[\frac{dy}{dx} = \frac{1}{\sqrt{3}} \text { at } t = \frac{2 \pi}{3}\] ?

 


Write the derivative of sinx with respect to cos x ?


Differentiate log (1 + x2) with respect to tan−1 x ?


If \[f\left( 1 \right) = 4, f'\left( 1 \right) = 2\] find the value of the derivative of  \[\log \left( f\left( e^x \right) \right)\] w.r. to x at the point x = 0 ?

 


If \[f\left( x \right) = \log \left\{ \frac{u \left( x \right)}{v \left( x \right)} \right\}, u \left( 1 \right) = v \left( 1 \right) \text{ and }u' \left( 1 \right) = v' \left( 1 \right) = 2\] , then find the value of `f' (1)` ?


If \[\sin \left( x + y \right) = \log \left( x + y \right), \text { then } \frac{dy}{dx} =\] ___________ .


If \[y = \tan^{- 1} \left( \frac{\sin x + \cos x}{\cos x - \sin x} \right), \text { then  } \frac{dy}{dx}\] is equal to ___________ .


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


If y = ex cos x, prove that \[\frac{d^2 y}{d x^2} = 2 e^x \cos \left( x + \frac{\pi}{2} \right)\] ?


If x = f(t) and y = g(t), then write the value of \[\frac{d^2 y}{d x^2}\] ?


If y = sin (m sin−1 x), then (1 − x2) y2 − xy1 is equal to


If y = (sin−1 x)2, then (1 − x2)y2 is equal to

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×