If y= √(1-x/1+x) then dy/dx equals:
Answers
Step-by-step explanation:
Answer:
(1-x^2)\frac{dy}{dx}+y=0(1−x
2
)
dx
dy
+y=0
Step-by-step explanation:
Given: y=\sqrt{\frac{1-x}{1+x}}y=
1+x
1−x
To prove: (1-x^2)\frac{dy}{dx}+y=0(1−x
2
)
dx
dy
+y=0
Proof:
The given equation is
y=\sqrt{\frac{1-x}{1+x}}y=
1+x
1−x
Squaring both sides we get
y^2=\frac{1-x}{1+x}y
2
=
1+x
1−x
Differential with respect to x.
2y\frac{dy}{dx}=\frac{(1+x)\frac{d}{dx}(1-x)-(1-x)\frac{d}{dx}(1+x)}{(1+x)^2}2y
dx
dy
=
(1+x)
2
(1+x)
dx
d
(1−x)−(1−x)
dx
d
(1+x)
[\because (\frac{f}{g})'=\frac{g\cdot f'-f\cdot g'}{g^2}][∵(
g
f
)
′
=
g
2
g⋅f
′
−f⋅g
′
]
2y\frac{dy}{dx}=\frac{(1+x)(-1)-(1-x)(1)}{(1+x)^2}2y
dx
dy
=
(1+x)
2
(1+x)(−1)−(1−x)(1)
2y\frac{dy}{dx}=\frac{-1-x-1+x}{(1+x)^2}2y
dx
dy
=
(1+x)
2
−1−x−1+x
2y\frac{dy}{dx}=\frac{-2}{(1+x)^2}2y
dx
dy
=
(1+x)
2
−2
Divide both sides by 2y.
\frac{dy}{dx}=\frac{\frac{-2}{(1+x)^2}}{2y}
dx
dy
=
2y
(1+x)
2
−2
\frac{dy}{dx}=-\frac{1}{y(1+x)^2}
dx
dy
=−
y(1+x)
2
1
We need to prove
(1-x^2)\frac{dy}{dx}+y=0(1−x
2
)
dx
dy
+y=0
Taking LHS,
LHS=(1-x^2)\frac{dy}{dx}+yLHS=(1−x
2
)
dx
dy
+y
Substitute the value of \frac{dy}{dx}
dx
dy
.
LHS=(1-x^2)(-\frac{1}{y(1+x)^2})+yLHS=(1−x
2
)(−
y(1+x)
2
1
)+y
LHS=-\frac{(1-x)(1+x)}{y(1+x)^2}+yLHS=−
y(1+x)
2
(1−x)(1+x)
+y
Cancel out the common factors.
LHS=-\frac{1-x}{y(1+x)}+yLHS=−
y(1+x)
1−x
+y
LHS=-\frac{1}{y}\times \frac{1-x}{(1+x)}+yLHS=−
y
1
×
(1+x)
1−x
+y
LHS=-\frac{1}{y}\times y^2+yLHS=−
y
1
×y
2
+y [\because y^2=\frac{1-x}{(1+x)}][∵y
2
=
(1+x)
1−x
]
Cancel out the common factors.
LHS=-y+yLHS=−y+y
LHS=0LHS=0