y=ln 1_cos2x÷1+cos2x ,find dy \dx
Answers
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