Question

If sin(x+y)=ycos(x+y) then prove that dy/dx=–(1+ysquare/ysquare)

Answer 1

Step-by-step explanation:

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Answer 2

$\underline{\textbf{\large{Given :}}}$

sin(x+y)=ycos(x+y)

$\underline{\textbf{\large{To prove:}}}$

$\frac{dy}{dx}=- (\frac{1 + {y}^{2}}{{y}^{2}})$

$\underline{\textbf{\large{Proof:}}}$

$\sin(x + y) = y\cos(x + y)$

$y = \frac{ \sin(x + y) }{ \cos(x + y) }$

$y = \tan(x + y)$

Differentiation wrt x

$( \frac{dy}{dx} ) = \frac{d}{dx} ( \tan(x + y) )$

$( \frac{dy}{dx} ) = { \sec}^{2} (x + y) \times (1 + \frac{dy}{dx} )$

$(\frac{dy}{dx} ) = { \sec}^{2} (x + y) + { \sec }^{2} (x + y)( \frac{dy}{dx} )$

$( \frac{dy}{dx}) - { \sec }^{2} ( x + y)( \frac{dy}{dx} ) = { \sec}^{2} (x + y)$

$(\frac{dy}{dx} )(1 - { \sec }^{2} (x + y)) = 1 + { \tan }^{2} (x + y)$

$(\frac{dy}{dx} )(1 - (1 + tan^2 (x + y)) = 1 + { \tan }^{2} (x + y)$

$(\frac{dy}{dx} )(1 - 1 - tan^2 (x + y)) = 1 + { \tan }^{2} (x + y)$

$(\frac{dy}{dx} )( - tan^2 (x + y)) = 1 + { \tan }^{2} (x + y)$

$\frac{dy}{dx} = \frac{(1 + { \tan}^{2}(x + y)) }{ - { \tan}^{2} (x + y)}$

$\frac{dy}{dx} = \frac{1 + {y}^{2} }{- {y}^{2} }$

$\frac{dy}{dx}=- (\frac{1 + {y}^{2}}{{y}^{2}})$