हिंदी

If Y = E X E X + X E E X + E X X E , Prove that D Y D X = E X E X ⋅ X E X { E X X + E X ⋅ Log X } - Mathematics

Advertisements
Advertisements

प्रश्न

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

 

योग
Advertisements

उत्तर

\[\text{ We have, y } = e^{x^{e^x}} + x^{e^{e^x}} + e^{x^{x^e}} \]
\[ \Rightarrow y = u + v + w\]
\[ \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} . . . \left( i \right)\]
\[\text{ where u } = e^{x^{e^x}} , v = x^{e^{e^x}} \text{ and w } = e^{x^{x^e}} \]
\[\text{ Now, u } = e^{x^{e^x}} . . . \left( ii \right)\]

Taking log on both sides,

\[\log u = \log e^{x^{e^x}} \]
\[ \Rightarrow \log u = x^{e^x} \log e\]
\[ \Rightarrow \log u = x^{e^x} . . . \left( iii \right)\]

Taking log on both sides,

\[\log \log u = \log x^{e^x} \]
\[ \Rightarrow \log \log u = e^x \log x\]

Differentiating with respect to x,

\[\Rightarrow \frac{1}{\log u}\frac{d}{dx}\left( \log u \right) = e^x \frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( e^x \right)\]
\[ \Rightarrow \frac{1}{\log u}\frac{1}{u}\frac{du}{dx} = \frac{e^x}{x} + e^x \log x\]
\[ \Rightarrow \frac{du}{dx} = u\log u\left[ \frac{e^x}{x} + e^x \log x \right]\]
\[ \Rightarrow \frac{du}{dx} = e^{x^{e^x}} \times x^{e^x} \left[ \frac{e^x}{x} + e^x \log x \right] . . . \left( A \right)\]
\[ \left[ \text{ Using equation } \left( ii \right) \text{ and } \left( iii \right) \right]\]
\[\text{ Now, v } = x^{e^{e^x}} . . . \left( iv \right)\]

Taking log on both sides,

\[\log v = \log x^{e^{e^x}} \]
\[ \Rightarrow \log v = e^{e^x} \log x\]

\[\Rightarrow \frac{1}{v}\frac{dv}{dx} = e^{e^x} \frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( e^{e^x} \right)\]
\[ \Rightarrow \frac{1}{v}\frac{dv}{dx} = e^{e^x} \left( \frac{1}{x} \right) + \log x e^{e^x} \frac{d}{dx}\left( e^x \right)\]
\[ \Rightarrow \frac{dv}{dx} = v\left[ e^{e^x} \left( \frac{1}{x} \right) + \log x e^{e^x} e^x \right]\]
\[ \Rightarrow \frac{dv}{dx} = x^{e^{e^x}} \times e^{e^x} \left[ \frac{1}{x} + e^x \log x \right] . . . \left( B \right) \]
\[ \left\{ \text{ Using equation } \left( 4 \right) \right\}\]
\[\text{ Now, w } = e^{x^{x^e}} . . . \left( v \right)\]

Taking log on both sides,

\[\log w = \log e^{x^{x^e}} \]
\[ \Rightarrow \log w = x^{x^e} \log e\]
\[ \Rightarrow \log w = x^{x^e} . . . \left( vi \right)\]

Taking log on both sides,

\[\log \log w = \log x^{x^e} \]
\[ \Rightarrow \log \log w = x^e \log x\]

\[\Rightarrow \frac{1}{\log w}\frac{d}{dx}\left( \log w \right) = x^e \frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( x^e \right)\]
\[ \Rightarrow \frac{1}{\log w}\left( \frac{1}{w} \right)\frac{dw}{dx} = x^e \left( \frac{1}{x} \right) + \log xe x^{e - 1} \]
\[ \Rightarrow \frac{dw}{dx} = w \log w\left[ x^{e - 1} + e \log x x^{e - 1} \right]\]
\[ \Rightarrow \frac{dw}{dx} = e^{x^{x^e}} x^{x^e} x^{e - 1} \left( 1 + e \log x \right) - - - - \left( C \right) \]
\[ \left[ \text{ using equation} \left( v \right), \left( vi \right) \right]\]
\[\text{ Using equation} \left( A \right), \left( B \right)\text{  and } \left( C \right) \text{ in equation } \left( i \right),\text{ we get }\]
\[\frac{dy}{dx} = e^{x^{e^x}} x^{e^x} \left[ \frac{e^x}{x} + e^x \log x \right] + x^{e^{e^x}} \times e^{e^x} \left[ \frac{1}{x} + e^x \log x \right] + e^{x^{x^e}} x^{x^e} x^{e - 1} \left( 1 + e \log x \right)\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Differentiation - Exercise 11.06 [पृष्ठ ९९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.06 | Q 7 | पृष्ठ ९९

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

 

If y = xx, prove that `(d^2y)/(dx^2)−1/y(dy/dx)^2−y/x=0.`

 

Differentiate the following functions from first principles log cos x ?


Differentiate the following functions from first principles x2ex ?


Differentiate \[3^{x^2 + 2x}\] ?


Differentiate \[\log \left( x + \sqrt{x^2 + 1} \right)\] ?


Differentiate \[e^{ax} \sec x \tan 2x\] ?


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


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


Differentiate \[\tan^{- 1} \left( \frac{2^{x + 1}}{1 - 4^x} \right), - \infty < x < 0\] ?


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


If  \[y = se c^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right), x > 0 . \text{ Find} \frac{dy}{dx}\] ?

 


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 the derivative of tan−1 (a + bx) takes the value 1 at x = 0, prove that 1 + a2 = b ?


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


If \[\cos y = x \cos \left( a + y \right), \text{ with } \cos a \neq \pm 1, \text{ prove that } \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\] ?


Differentiate \[{10}^{ \log \sin x }\] ?


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

 


If `y=(sinx)^x + sin^-1 sqrtx  "then find"  dy/dx` 


\[y = \left( \sin x \right)^{\left( \sin x \right)^{\left( \sin x \right)^{. . . \infty}}} \],prove that \[\frac{y^2 \cot x}{\left( 1 - y \log \sin x \right)}\] ?


If  \[x = a\sin2t\left( 1 + \cos2t \right) \text { and y } = b\cos2t\left( 1 - \cos2t \right)\] , show that at  \[t = \frac{\pi}{4}, \frac{dy}{dx} = \frac{b}{a}\] ?


\[\text { If }x = \cos t\left( 3 - 2 \cos^2 t \right), y = \sin t\left( 3 - 2 \sin^2 t \right) \text { find the value of } \frac{dy}{dx}\text{ at }t = \frac{\pi}{4}\] ?


If  \[x = \frac{1 + \log t}{t^2}, y = \frac{3 + 2\log t}{t}, \text { find } \frac{dy}{dx}\] ?


Differentiate (log x)x with respect to log x ?


\[\sin^{- 1} \sqrt{1 - x^2}\] with respect to \[\cot^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right),\text { if }0 < x < 1\] ? 


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 sin (log x) ?


If y = x3 log x, prove that \[\frac{d^4 y}{d x^4} = \frac{6}{x}\] ?


If \[y = \frac{\log x}{x}\] show that \[\frac{d^2 y}{d x^2} = \frac{2 \log x - 3}{x^3}\] ?


If x = a(1 − cos θ), y = a(θ + sin θ), prove that \[\frac{d^2 y}{d x^2} = - \frac{1}{a}\text { at } \theta = \frac{\pi}{2}\] ?


If y = (sin−1 x)2, prove that (1 − x2)

\[\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] ?


If \[y = e^{2x} \left( ax + b \right)\]  show that  \[y_2 - 4 y_1 + 4y = 0\] ?


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 \[\frac{d^2 y}{d x^2}\text{ at } t = \frac{\pi}{2}\] ?


\[\text { If x } = \cos t + \log \tan\frac{t}{2}, y = \sin t, \text { then find the value of } \frac{d^2 y}{d t^2} \text { and } \frac{d^2 y}{d x^2} \text { at } t = \frac{\pi}{4} \] ?


\[\text { Find A and B so that y = A } \sin3x + B \cos3x \text { satisfies the equation }\]

\[\frac{d^2 y}{d x^2} + 4\frac{d y}{d x} + 3y = 10 \cos3x \] ?


\[\text { If y } = x^n \left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\}, \text { prove that } x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)x\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0 \] 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 \[y = \log_e \left( \frac{x}{a + bx} \right)^x\] then x3 y2 =

 


Differentiate the following with respect to x

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


f(x) = xx has a stationary point at ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×