Math, asked by mohapatraleena4, 3 months ago

y=ln 1_cos2x÷1+cos2x ,find dy \dx

Answers

Answered by suvarnaj728
0

Step-by-step explanation:

=

1−cos2x

1+cos2x

\textbf{To find:}To find:

\mathsf{\dfrac{dy}{dx}}

dx

dy

\textbf{Solution:}Solution:

\textsf{Consider,}Consider,

\mathsf{y=\dfrac{\sqrt{1+cos\,2x}}{\sqrt{1-cos\,2x}}}y=

1−cos2x

1+cos2x

\mathsf{Using,}Using,

\boxed{\mathsf{cos2x=1-2\,sin^2x\;\implies\;1-cos2x=2\,sin^2x}}

cos2x=1−2sin

2

x⟹1−cos2x=2sin

2

x

\boxed{\mathsf{cos2x=2\,cos^2x-1\;\implies\;1+cos2x=2\,cos^2x}}

cos2x=2cos

2

x−1⟹1+cos2x=2cos

2

x

\mathsf{y=\dfrac{\sqrt{2\,cos^2x}}{\sqrt{2\,sin^2x}}}y=

2sin

2

x

2cos

2

x

\mathsf{y=\dfrac{\sqrt{2}\,cosx}{\sqrt{2}\,sinx}}}

\mathsf{y=\dfrac{cosx}{sinx}}y=

sinx

cosx

\mathsf{y=cotx}y=cotx

\textsf{Differentiate with respect to 'x'}Differentiate with respect to ’x’

\boxed{\mathsf{\dfrac{dy}{dx}=-cosec^2x}}

dx

dy

=−cosec

2

x

\textbf{Find more:}Find more:

If y=e^2x (ax+b), then prove that d^2 y/dx^2 -4 dy/dx+4y=0

https://brainly.in/question/19674230

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