write short note on environment ?
Answers
Answer:
Environment is the nature and surroundings in which all plants, animals, humans and other living beings live and operate. ... It includes sunlight, atmosphere, land, water, plants, animals, sea life, minerals, different species and everything that occurs naturally on earth.
Answer:
\large\underline{\sf{Solution-}}
Solution−
Given to differentiate
\begin{gathered}\rm \: {sin}^{ - 1}(2x \sqrt{1 - {x}^{2} }) \: w.r.t \: \: {sin}^{ - 1}x \\ \end{gathered}
sin
−1
(2x
1−x
2
)w.r.tsin
−1
x
Let assume that
\begin{gathered}\rm \: u \: = \: {sin}^{ - 1}(2x \sqrt{1 - {x}^{2} }) \: \\ \end{gathered}
u=sin
−1
(2x
1−x
2
)
and
\begin{gathered}\rm \: v = {sin}^{ - 1}x \\ \end{gathered}
v=sin
−1
x
Now, Consider
\begin{gathered}\rm \: u \: = \: {sin}^{ - 1}(2x \sqrt{1 - {x}^{2} }) \: \\ \end{gathered}
u=sin
−1
(2x
1−x
2
)
Let we substitute,
\begin{gathered}\rm \: x = sin \theta \: \\ \end{gathered}
x=sinθ
So, above expression can be rewritten as
\begin{gathered}\rm \: u = {sin}^{ - 1}(2sin\theta \sqrt{1 - {sin}^{2}\theta } ) \\ \end{gathered}
u=sin
−1
(2sinθ
1−sin
2
θ
)
\begin{gathered}\rm \: u = {sin}^{ - 1}(2sin\theta \sqrt{{cos}^{2}\theta } ) \\ \end{gathered}
u=sin
−1
(2sinθ
cos
2
θ
)
\begin{gathered}\rm \: u = {sin}^{ - 1}(2sin\theta \: cos\theta ) \\ \end{gathered}
u=sin
−1
(2sinθcosθ)
\begin{gathered}\rm \: u = {sin}^{ - 1}(sin2\theta ) \\ \end{gathered}
u=sin
−1
(sin2θ)
Given that,
\begin{gathered}\rm \: \dfrac{1}{ \sqrt{2} } < x < 1 \\ \end{gathered}
2
1
<x<1
\begin{gathered}\rm \: \dfrac{1}{ \sqrt{2} } < sin\theta < 1 \\ \end{gathered}
2
1
<sinθ<1
\begin{gathered}\rm\implies \:\dfrac{\pi}{4} < \theta < \dfrac{\pi}{2} \\ \end{gathered}
⟹
4
π
<θ<
2
π
\begin{gathered}\rm\implies \:\dfrac{\pi}{2} < 2 \theta < \pi \\ \end{gathered}
⟹
2
π
<2θ<π
\begin{gathered}\rm\implies \: - \pi < - 2 \theta < - \dfrac{\pi}{2} \\ \end{gathered}
⟹−π<−2θ<−
2
π
\begin{gathered}\rm\implies \: \pi- \pi < \pi - 2 \theta < \pi - \dfrac{\pi}{2} \\ \end{gathered}
⟹π−π<π−2θ<π−
2
π
\begin{gathered}\rm\implies \: 0 < \pi - 2 \theta < \dfrac{\pi}{2} \\ \end{gathered}
⟹0<π−2θ<
2
π
So, above expression
\begin{gathered}\rm \: u = {sin}^{ - 1}(sin2\theta ) \\ \end{gathered}
u=sin
−1
(sin2θ)
can be rewritten as
\begin{gathered}\rm \: u = {sin}^{ - 1}[sin(\pi - 2\theta )] \\ \end{gathered}
u=sin
−1
[sin(π−2θ)]
We know,
\begin{gathered}\boxed{\sf{ \:{sin}^{ - 1}(sinx) = x \: \: if -\dfrac{\pi }{2} \leqslant x \leqslant \dfrac{\pi }{2} \: }} \\ \end{gathered}
sin
−1
(sinx)= x if−
2
π
⩽x⩽
2
π
So, using this, we get
\begin{gathered}\rm \: u = \pi - 2\theta \\ \end{gathered}
u=π−2θ
\begin{gathered}\rm \: u = \pi - 2{sin}^{ - 1}x \\ \end{gathered}
u=π−2sin
−1
x
\begin{gathered}\rm \: [ \: \because \: x = sin\theta \: \rm\implies \:\theta = {sin}^{ - 1}x \: ] \\ \end{gathered}
[∵x=sinθ⟹θ=sin
−1
x]
On differentiating w. r. t. x, we get
\begin{gathered}\rm \: \dfrac{d}{dx}u \: = \: \dfrac{d}{dx}(\pi - 2{sin}^{ - 1}x) \\ \end{gathered}
dx
d
u=
dx
d
(π−2sin
−1
x)
\begin{gathered}\rm \: \dfrac{du}{dx} \: = \: 0 - \dfrac{2}{ \sqrt{1 - {x}^{2} } } \\ \end{gathered}
dx
du
=0−
1−x
2
2
\begin{gathered}\rm\implies \:\boxed{\sf{ \: \: \rm \: \dfrac{du}{dx} \: = \: - \dfrac{2}{ \sqrt{1 - {x}^{2} } } \: }} \\ \end{gathered}
⟹
dx
du
=−
1−x
2
2
Now, Consider
\begin{gathered}\rm \:v \: = \: {sin}^{ - 1}x \\ \end{gathered}
v=sin
−1
x
On differentiating both sides w. r. t. x, we get
\begin{gathered}\rm \: \dfrac{d}{dx}v \: = \: \dfrac{d}{dx}{sin}^{ - 1}x \\ \end{gathered}
dx
d
v=
dx
d
sin
−1
x
\begin{gathered}\rm \: \dfrac{dv}{dx} \: = \: \dfrac{1}{ \sqrt{1 - {x}^{2} } } \\ \end{gathered}
dx
dv
=
1−x
2
1
Now, Consider
\begin{gathered}\rm \: \dfrac{du}{dv} \\ \end{gathered}
dv
du
\begin{gathered}\rm \: = \: \dfrac{du}{dx} \div \dfrac{dv}{dx} \\ \end{gathered}
=
dx
du
÷
dx
dv
\begin{gathered}\rm \: = \: - \: \dfrac{2}{ \sqrt{1 - {x}^{2} } } \div \dfrac{1}{ \sqrt{1 - {x}^{2} } } \\ \end{gathered}
=−
1−x
2
2
÷
1−x
2
1
\begin{gathered}\rm \: = \: - 2 \\ \end{gathered}
=−2
Hence,
\begin{gathered}\rm\implies \:\rm \: \dfrac{du}{dv} \: = \: - \: 2 \\ \end{gathered}
⟹
dv
du
=−2
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
Additional Information :-
\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = {sin}^{ - 1}(sinx) & \sf x \: \: if -\dfrac{\pi }{2} \leqslant x \leqslant \dfrac{\pi }{2}\\ \\ \sf y = {cos}^{ - 1}(cosx) & \sf x \: \: if \: 0 \leqslant y \leqslant \pi \\ \\ \sf y = {tan}^{ - 1}(tanx) & \sf x \: \: if \: - \dfrac{\pi }{2} < x < \dfrac{\pi }{2}\\ \\ \sf y = {cosec}^{ - 1}(cosecx) & \sf x \: \: if \: x \: \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi }{2}\bigg] - \{0 \}\\ \\ \sf y = {sec}^{ - 1}(secx) & \sf x \: \: if \: x \: \in \: [0, \: \pi] \: - \: \bigg\{\dfrac{\pi }{2}\bigg\}\\ \\ \sf y = {cot}^{ - 1}(cotx) & \sf x \: \: if \: \: \in \: \bigg( - \dfrac{\pi }{2} , \dfrac{\pi }{2}\bigg) - \{0 \} \end{array}} \\ \end{gathered} \\\end{gathered}
Function
y= sin
−1
(sinx)
y= cos
−1
y= (cosx)
y= tan
−1
(tanx)
y= cosec
−1
(cosecx)
sec