Advertisements
Advertisements
Question
If \[\sin y = x \sin \left( a + y \right), \text { then }\frac{dy}{dx} \text { is}\] ____________ .
Options
\[\frac{\sin a}{\sin a \sin^2 \left( a + y \right)}\]
\[\frac{\sin^2 \left( a + y \right)}{\sin a}\]
\[\sin a \sin^2 \left( a + y \right)\]
\[\frac{\sin^2 \left( a - y \right)}{\sin a}\]
Solution
\[\frac{\sin^2 \left( a + y \right)}{\sin a}\]
\[\text { We have,} \sin y = x \sin\left( a + y \right)\]
\[\Rightarrow \frac{d}{dx}\left( \sin y \right) = \frac{d}{dx}\left[ x \sin\left( a + y \right) \right]\]
\[ \Rightarrow \cos y\frac{dy}{dx} = \sin\left( a + y \right)\frac{d}{dx}\left( x \right) + x\frac{d}{dx}\left\{ \sin\left( a + y \right) \right\}\]
\[ \Rightarrow \cos y\frac{dy}{dx} = \sin\left( a + y \right) \times 1 + x \cos\left( a + y \right)\frac{dy}{dx}\]
\[ \Rightarrow \cos y\frac{dy}{dx} = \sin\left( a + y \right) + x \cos\left( a + y \right)\frac{dy}{dx}\]
\[ \Rightarrow \cos y\frac{dy}{dx} - x \cos\left( a + y \right)\frac{dy}{dx} = \sin\left( a + y \right)\]
\[ \Rightarrow \left\{ \cos y - x \cos \left( a + y \right) \right\}\frac{dy}{dx} = \sin\left( a + y \right)\]
\[ \Rightarrow \left\{ \cos y - \frac{\sin y}{\sin\left( a + y \right)} \times \cos\left( a + y \right) \right\}\frac{dy}{dx} = \sin\left( a + y \right) .............\binom{\because \sin y = 2 \sin x \cos x}{ \therefore x = \frac{\sin y}{\sin\left( a + y \right)}}\]
\[ \Rightarrow \left\{ \frac{\sin\left( a + y \right) \cos y - \sin y \cos\left( a + y \right)}{\sin\left( a + y \right)} \right\}\frac{dy}{dx} = \sin\left( a + y \right)\]
\[ \Rightarrow \frac{\sin\left( a + y - y \right)}{\sin\left( a + y \right)} \times \frac{dy}{dx} = \sin\left( a + y \right) \]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a}\]
APPEARS IN
RELATED QUESTIONS
Differentiate \[3^{e^x}\] ?
Differentiate \[\sqrt{\frac{1 + x}{1 - x}}\] ?
Differentiate \[\frac{2^x \cos x}{\left( x^2 + 3 \right)^2}\] ?
Differentiate \[\log \sqrt{\frac{x - 1}{x + 1}}\] ?
If \[y = \frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)\] , prove that \[\frac{ dy }{ dx } = 2 \text{cosec }2x \] ?
Differentiate \[\sin^{- 1} \left( 2 x^2 - 1 \right), 0 < x < 1\] ?
Differentiate \[\sin^{- 1} \left( 1 - 2 x^2 \right), 0 < x < 1\] ?
Differentiate \[\sin^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?
Differentiate \[\tan^{- 1} \left( \frac{2^{x + 1}}{1 - 4^x} \right), - \infty < x < 0\] ?
Differentiate \[\sin^{- 1} \left( \frac{1}{\sqrt{1 + x^2}} \right)\] ?
Differentiate
\[\tan^{- 1} \left( \frac{\cos x + \sin x}{\cos x - \sin x} \right), \frac{\pi}{4} < x < \frac{\pi}{4}\] ?
If \[y = \sin^{- 1} \left( 6x\sqrt{1 - 9 x^2} \right), - \frac{1}{3\sqrt{2}} < x < \frac{1}{3\sqrt{2}}\] \[\frac{dy}{dx} \] ?
Find \[\frac{dy}{dx}\] in the following case \[xy = c^2\] ?
Find \[\frac{dy}{dx}\] in the following case \[4x + 3y = \log \left( 4x - 3y \right)\] ?
If \[\sin \left( xy \right) + \frac{y}{x} = x^2 - y^2 , \text{ find} \frac{dy}{dx}\] ?
If \[e^x + e^y = e^{x + y} , \text{ prove that } \frac{dy}{dx} = - \frac{e^x \left( e^y - 1 \right)}{e^y \left( e^x - 1 \right)} or \frac{dy}{dx} + e^{y - x} = 0\] ?
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 \[\left( 1 + \cos x \right)^x\] ?
Differentiate \[x^{\tan^{- 1} x }\] ?
Differentiate \[x^{x \cos x +} \frac{x^2 + 1}{x^2 - 1}\] ?
Differentiate\[\left( x + \frac{1}{x} \right)^x + x^\left( 1 + \frac{1}{x} \right)\] ?
If \[y = x \sin \left( a + y \right)\] , prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin \left( a + y \right) - y \cos \left( a + y \right)}\] ?
If \[x = a \left( \frac{1 + t^2}{1 - t^2} \right) \text { and y } = \frac{2t}{1 - t^2}, \text { find } \frac{dy}{dx}\] ?
If f (x) = loge (loge x), then write the value of `f' (e)` ?
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 \[f\left( x \right) = \sqrt{x^2 + 6x + 9}, \text { then } f'\left( x \right)\] is equal to ______________ .
Find the second order derivatives of the following function sin (log x) ?
Find the second order derivatives of the following function log (sin x) ?
Find the second order derivatives of the following function x cos x ?
If x = a (1 − cos3 θ), y = a sin3 θ, prove that \[\frac{d^2 y}{d x^2} = \frac{32}{27a} \text { at } \theta = \frac{\pi}{6}\] ?
If x = a (θ − sin θ), y = a (1 + cos θ) prove that, find \[\frac{d^2 y}{d x^2}\] ?
Find \[\frac{d^2 y}{d x^2}\] where \[y = \log \left( \frac{x^2}{e^2} \right)\] ?
If x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, find \[\frac{d^2 y}{d x^2}\text{ at } t = \frac{\pi}{2}\] ?
\[\text { If y } = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} , \text { prove that }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \]
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
\[\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \] ?
If \[y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!}\] .....to ∞, then write \[\frac{d^2 y}{d x^2}\] in terms of y ?
If y = axn+1 + bx−n, then \[x^2 \frac{d^2 y}{d x^2} =\]
If y = a sin mx + b cos mx, then \[\frac{d^2 y}{d x^2}\] is equal to
If y2 = ax2 + bx + c, then \[y^3 \frac{d^2 y}{d x^2}\] is
If y = xx, prove that \[\frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 .\]
