Question

निम्नलिखित फलनों के अवकलन ज्ञात कीजिए (यह समझा जाय कि a, b, c, d,p, q, r और s निशिचत शून्येतर अचर हैं m तथा n पूर्णांक हैंl ) : $\dfrac{\sin x + \cos x}{\sin x - \cos x}$

Answer 1

Answer:

Step-by-step explanation:

माना कि  

           $y=\dfrac{\sin x + \cos x}{\sin x - \cos x}$

$\dfrac{dy}{dx} =\dfrac{(sinx-cosx)\dfrac{d}{dx} (sinx+cosx)-(sinx+cosx)\dfrac{d}{dx}(sinx-cosx)}{(sinx-cosx)^2}\\ \\=\dfrac{(sinx-cosx)(cosx-sinx)-(sinx+cosx)(cosx+sinx)}{(sinx-cosx)^2}\\\\=\dfrac{-(cosx-sinx)^2-(sinx+cosx)^2}{(sinx-cosx)^2}\\\\=\dfrac{-(cos^2x+sin^2x-2cosxsinx)-(cos^2x+sin^2x+2sinxcosx)}{(sinx-cosx)^2}\\\\=\dfrac{-(1-2sinxcosx)-(1+2sinxcosx)}{(sinx-cosx)^2}\\\\=\dfrac{-1+2sinxcosx-1-2sinxcosx}{(sinx-cosx)^2}\\\\=\dfrac{-2}{(sinx-cosx)^2}$

Answer 2

Answer:

Answer:

Step-by-step explanation:

माना कि

y=\dfrac{\sin x + \cos x}{\sin x - \cos x}y=

sinx−cosx

sinx+cosx

\begin{gathered}\dfrac{dy}{dx} =\dfrac{(sinx-cosx)\dfrac{d}{dx} (sinx+cosx)-(sinx+cosx)\dfrac{d}{dx}(sinx-cosx)}{(sinx-cosx)^2}\\ \\=\dfrac{(sinx-cosx)(cosx-sinx)-(sinx+cosx)(cosx+sinx)}{(sinx-cosx)^2}\\\\=\dfrac{-(cosx-sinx)^2-(sinx+cosx)^2}{(sinx-cosx)^2}\\\\=\dfrac{-(cos^2x+sin^2x-2cosxsinx)-(cos^2x+sin^2x+2sinxcosx)}{(sinx-cosx)^2}\\\\=\dfrac{-(1-2sinxcosx)-(1+2sinxcosx)}{(sinx-cosx)^2}\\\\=\dfrac{-1+2sinxcosx-1-2sinxcosx}{(sinx-cosx)^2}\\\\=\dfrac{-2}{(sinx-cosx)^2}\end{gathered}

dx

dy

=

(sinx−cosx)

2

(sinx−cosx)

dx

d

(sinx+cosx)−(sinx+cosx)

dx

d

(sinx−cosx)

=

(sinx−cosx)

2

(sinx−cosx)(cosx−sinx)−(sinx+cosx)(cosx+sinx)

=

(sinx−cosx)

2

−(cosx−sinx)

2

−(sinx+cosx)

2

=

(sinx−cosx)

2

−(cos

2

x+sin

2

x−2cosxsinx)−(cos

2

x+sin

2

x+2sinxcosx)

=

(sinx−cosx)

2

−(1−2sinxcosx)−(1+2sinxcosx)

=

(sinx−cosx)

2

−1+2sinxcosx−1−2sinxcosx

=

(sinx−cosx)

2

−2