मराठी

Differentiate the Following Functions from First Principles Ecos X. - Mathematics

Advertisements
Advertisements

प्रश्न

Differentiate the following functions from first principles ecos x.

Advertisements

उत्तर

\[\text{Let } f \left( x \right) = e^{\cos x} \]
\[ \Rightarrow f\left( x + h \right) = e^{\cos\left( x + h \right)} \]
\[\therefore \frac{d}{dx}\left\{ f\left( x \right) \right\} = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{e^{\cos\left( x + h \right)} - e^{\cos x}}{h}\]
\[ = \lim_{h \to 0} e^{\cos x }\left[ \frac{e^{\cos\left( x + h \right) - \cos x} - 1}{h} \right]\]
\[ = \lim_{h \to 0} e^{\cos x} \left[ \frac{e^{\cos\left( x + h \right) - \cos x} - 1}{\cos\left( x + h \right) \cos x} \right] \times \frac{\cos\left( x + h \right) - \ cosx}{h}\]
\[ = e^{\cos x} \lim_{h \to 0} \left( \frac{cos\left( x + h \right) - \cos x}{h} \right) \times \lim_{h \to 0} \left[ \frac{e^{\cos\left( x + h \right) - \ cos x} - 1}{\cos\left( x + h \right) - \cos x} \right]\]
\[ = e^{\cos x} \lim_{h \to 0} \left( \frac{\cos\left( x + h \right) - \cos x}{h} \right) \left[ \because \lim_{h \to 0} \frac{e^x - 1}{x} = 1 \right]\]
\[ = e^{\cos x} \lim_{h \to 0} \left\{ \frac{- 2\sin\left( \frac{x + h + x}{2} \right)\sin\left( \frac{x + h - x}{2} \right)}{h} \right\} \left[ \because \cos A - \cos B = - 2\sin \left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right) \right]\]
\[ = e^{\cos x} \lim_{h \to 0} \frac{- \sin\left( \frac{2x + h}{2} \right)}{1} \times \frac{\sin\left( \frac{h}{2} \right)}{\frac{h}{2}}\]
\[ = e^{\cos x} \lim_{h \to 0} \frac{- \sin\left( \frac{2x + h}{2} \right)}{1} \times \lim_{h \to 0} \frac{\sin\left( \frac{h}{2} \right)}{\frac{h}{2}}\]
\[ = e^{\cos x} \lim_{h \to 0} - \sin\left( \frac{2x + h}{2} \right) \left[ \because \frac{\sin x}{x} = 1 \right]\]
\[ = e^{\cos x} \left( - \sin x \right)\]
\[ = - \sin x e^{\cos x} \]
\[\text{ Hence }, \frac{d}{dx}\left( e^{\cos x} \right) = - \sin x e^{\cos x }\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Differentiation - Exercise 11.01 [पृष्ठ १७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.01 | Q 4 | पृष्ठ १७

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Differentiate \[3^{x \log x}\] ?


Differentiate \[\left( \sin^{- 1} x^4 \right)^4\] ?


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


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


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


Differentiate \[\sin^{- 1} \left\{ \frac{\sin x + \cos x}{\sqrt{2}} \right\}, - \frac{3 \pi}{4} < x < \frac{\pi}{4}\] ?


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


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


Differentiate \[\sin^{- 1} \left( \frac{1}{\sqrt{1 + x^2}} \right)\] with respect to x.


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


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


If \[y = x \sin y\] , Prove that \[\frac{dy}{dx} = \frac{\sin y}{\left( 1 - x \cos y \right)}\] ?


If \[e^x + e^y = e^{x + y} , \text{ prove that } \frac{dy}{dx} = - \frac{e^x \left( e^y - 1 \right)}{e^y \left( e^x - 1 \right)} or \frac{dy}{dx} + e^{y - x} = 0\] ?


Find  \[\frac{dy}{dx}\] \[y = e^x + {10}^x + x^x\] ?

 


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


If \[e^y = y^x ,\] prove that\[\frac{dy}{dx} = \frac{\left( \log y \right)^2}{\log y - 1}\] ?


If  \[y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + . . to \infty}}}\] , prove that \[\frac{dy}{dx} = \frac{\sec^2 x}{2 y - 1}\] ?

 


If \[\frac{dy}{dx}\] when \[x = a \cos \theta \text{ and } y = b \sin \theta\] ?


Find \[\frac{dy}{dx}\] , when \[x = b   \sin^2   \theta  \text{ and }  y = a   \cos^2   \theta\] ?


Find \[\frac{dy}{dx}\] , when  \[x = \cos^{- 1} \frac{1}{\sqrt{1 + t^2}} \text{ and y } = \sin^{- 1} \frac{t}{\sqrt{1 + t^2}}, t \in R\] ?


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

 


Differentiate \[\tan^{- 1} \left( \frac{\cos x}{1 + \sin x} \right)\] with  respect to \[\sec^{- 1} x\] ?


Differentiate \[\cos^{- 1} \left( 4 x^3 - 3x \right)\] with respect to \[\tan^{- 1} \left( \frac{\sqrt{1 - x^2}}{x} \right), \text{ if }\frac{1}{2} < x < 1\] ? 


If \[f'\left( 1 \right) = 2 \text { and y } = f \left( \log_e x \right), \text { find} \frac{dy}{dx} \text { at }x = e\] ?


Let g (x) be the inverse of an invertible function f (x) which is derivable at x = 3. If f (3) = 9 and `f' (3) = 9`, write the value of `g' (9)`.


If \[y = \log \sqrt{\tan x}, \text{ write } \frac{dy}{dx} \] ?


Given  \[f\left( x \right) = 4 x^8 , \text { then }\] _________________ .


If \[\sin^{- 1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = \text { log a then } \frac{dy}{dx}\] is equal to _____________ .


If y = x + tan x, show that  \[\cos^2 x\frac{d^2 y}{d x^2} - 2y + 2x = 0\] ?


If \[y = e^{\tan^{- 1} x}\] prove that (1 + x2)y2 + (2x − 1)y1 = 0 ?


If y = cot x show that \[\frac{d^2 y}{d x^2} + 2y\frac{dy}{dx} = 0\] ?


If x = 2 cos t − cos 2ty = 2 sin t − sin 2t, find \[\frac{d^2 y}{d x^2}\text{ at } t = \frac{\pi}{2}\] ?


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


If y = (cot−1 x)2, prove that y2(x2 + 1)2 + 2x (x2 + 1) y1 = 2 ?


If y = x + ex, find \[\frac{d^2 x}{d y^2}\] ?


\[\frac{d^{20}}{d x^{20}} \left( 2 \cos x \cos 3 x \right) =\]

 


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


If y = xx, prove that \[\frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 .\]


Differentiate the following with respect to x

\[\cot^{- 1} \left( \frac{1 - x}{1 + x} \right)\]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×