Advertisements
Advertisements
प्रश्न
If \[x = 2 \cos \theta - \cos 2 \theta \text{ and y} = 2 \sin \theta - \sin 2 \theta\], prove that \[\frac{dy}{dx} = \tan \left( \frac{3 \theta}{2} \right)\] ?
Advertisements
उत्तर
\[\Rightarrow \frac{dx}{d\theta} = 2\left( - \sin\theta \right) - \left( - \sin2\theta \right)\frac{d}{d\theta}\left( 2\theta \right)\]
\[ \Rightarrow \frac{dx}{d\theta} = - 2\sin\theta + 2 \sin2\theta\]
\[ \Rightarrow \frac{dx}{d\theta} = 2\left( \sin2\theta - \sin\theta \right) . . . \left( i \right)\]
\[\text{ and }, \]
\[y = 2 \sin\theta - \sin2\theta\]
\[\Rightarrow \frac{dy}{d\theta} = 2 \cos\theta - \cos2\theta\frac{d}{d\theta}\left( 2\theta \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = 2 \cos\theta - \cos2\theta\left( 2 \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = 2 \cos\theta - 2 \cos2\theta\]
\[ \Rightarrow \frac{dy}{d\theta} = 2\left( \cos\theta - \cos2\theta \right) . . . \left( ii \right)\]
\[\text{ Dividing equation } \left( ii \right) \text{ by equation } \left( i \right), \]
\[\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2\left( \cos\theta - \cos2\theta \right)}{2\left( \sin2\theta - \sin\theta \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\cos\theta - \cos2\theta}{\sin2\theta - \sin\theta}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 2\sin\left( \frac{\theta + 2\theta}{2} \right)\sin\left( \frac{\theta - 2\theta}{2} \right)}{2\cos\left( \frac{2\theta + \theta}{2} \right)\sin\left( \frac{2\theta - \theta}{2} \right)} ............[\because \sin A - \sin B = 2 \cos\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right)\text{ and } \cos A - \cos B = - 2\sin\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right)]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- \sin\left( \frac{3\theta}{2} \right)\sin\left( \frac{- \theta}{2} \right)}{\cos\left( \frac{3\theta}{2} \right)\sin\left( \frac{\theta}{2} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- \sin\left( \frac{3\theta}{2} \right)\left( - \sin\frac{\theta}{2} \right)}{\cos\left( \frac{3\theta}{2} \right)\sin\left( \frac{\theta}{2} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin\left( \frac{3\theta}{2} \right)}{\cos\left( \frac{3\theta}{2} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \tan\left( \frac{3\theta}{2} \right)\]
APPEARS IN
संबंधित प्रश्न
Differentiate the following functions from first principles \[e^\sqrt{2x}\].
Differentiate the following functions from first principles sin−1 (2x + 3) ?
Differentiate \[3^{x^2 + 2x}\] ?
Differentiate \[\sqrt{\frac{1 - x^2}{1 + x^2}}\] ?
Differentiate (log sin x)2 ?
If \[y = \frac{x}{x + 2}\] , prove tha \[x\frac{dy}{dx} = \left( 1 - y \right) y\] ?
If \[y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}\] , prove that \[\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}\] ?
If \[y = e^x \cos x\] ,prove that \[\frac{dy}{dx} = \sqrt{2} e^x \cdot \cos \left( x + \frac{\pi}{4} \right)\] ?
Differentiate \[\tan^{- 1} \left\{ \frac{x}{a + \sqrt{a^2 - x^2}} \right\}, - a < x < a\] ?
Differentiate \[\cos^{- 1} \left( \frac{1 - x^{2n}}{1 + x^{2n}} \right), < x < \infty\] ?
Differentiate \[\sin^{- 1} \left\{ \frac{2^{x + 1} \cdot 3^x}{1 + \left(36 \right)^x} \right\}\] with respect to x.
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 \[\cos y = x \cos \left( a + y \right), \text{ with } \cos a \neq \pm 1, \text{ prove that } \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\] ?
Differentiate \[x^{1/x}\] with respect to x.
Differentiate \[{10}^\left( {10}^x \right)\] ?
Differentiate \[\sin \left( x^x \right)\] ?
Differentiate \[x^{\tan^{- 1} x }\] ?
Differentiate \[\left( x \cos x \right)^x + \left( x \sin x \right)^{1/x}\] ?
Differentiate\[\left( x + \frac{1}{x} \right)^x + x^\left( 1 + \frac{1}{x} \right)\] ?
Find \[\frac{dy}{dx}\] \[y = \left( \tan x \right)^{\log x} + \cos^2 \left( \frac{\pi}{4} \right)\] ?
Find \[\frac{dy}{dx}\] \[y = x^{\log x }+ \left( \log x \right)^x\] ?
If \[x^m y^n = 1\] , prove that \[\frac{dy}{dx} = - \frac{my}{nx}\] ?
If \[y = \left( \tan x \right)^{\left( \tan x \right)^{\left( \tan x \right)^{. . . \infty}}}\], prove that \[\frac{dy}{dx} = 2\ at\ x = \frac{\pi}{4}\] ?
If \[x = \cos t \text{ and y } = \sin t,\] prove that \[\frac{dy}{dx} = \frac{1}{\sqrt{3}} \text { at } t = \frac{2 \pi}{3}\] ?
Differentiate \[\sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] with respect to \[\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right), \text { if } 0 < x < 1\] ?
The differential coefficient of f (log x) w.r.t. x, where f (x) = log x is ___________ .
The derivative of the function \[\cot^{- 1} \left| \left( \cos 2 x \right)^{1/2} \right| \text{ at } x = \pi/6 \text{ is }\] ______ .
If \[y = \frac{1}{1 + x^{a - b} +^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}}\] then \[\frac{dy}{dx}\] is equal to ______________ .
Find the second order derivatives of the following function e6x cos 3x ?
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 \[y = e^{\tan^{- 1} x}\] prove that (1 + x2)y2 + (2x − 1)y1 = 0 ?
If y = 3 cos (log x) + 4 sin (log x), prove that x2y2 + xy1 + y = 0 ?
\[\text { If x } = a\left( \cos t + t \sin t \right) \text { and y} = a\left( \sin t - t \cos t \right),\text { then find the value of } \frac{d^2 y}{d x^2} \text { at } t = \frac{\pi}{4} \] ?
\[\text { If y } = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} , \text { prove that }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \]
Disclaimer: There is a misprint in the question,
\[\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0\] must be written instead of
\[\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \] ?
Let f(x) be a polynomial. Then, the second order derivative of f(ex) is
If y = sin (m sin−1 x), then (1 − x2) y2 − xy1 is equal to
If \[y = \frac{ax + b}{x^2 + c}\] then (2xy1 + y)y3 =
\[\text { If } y = \left( x + \sqrt{1 + x^2} \right)^n , \text { then show that }\]
\[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = n^2 y .\]
