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Differentiation
JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation
JEE Main · Mathematics · Differentiation
If y= x+ y+ x+ y+ ∞ , then (d y)/(d x)=
If
y
=
x
+
y
+
x
+
y
+
…
∞
y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \infty}}}}
y
=
x
+
y
+
x
+
y
+
…
∞
, then
d
y
d
x
=
\frac{d y}{d x}=
d
x
d
y
=
Options
A
y
3
−
x
2
y
2
−
2
x
y
+
1
\frac{y^3-x}{2 y^2-2 x y+1}
2
y
2
−
2
x
y
+
1
y
3
−
x
B
x
+
y
3
2
y
2
−
x
\frac{x+y^3}{2 y^2-x}
2
y
2
−
x
x
+
y
3
C
y
+
x
y
2
−
2
x
\frac{y+x}{y^2-2 x}
y
2
−
2
x
y
+
x
D
y
2
−
x
2
y
3
−
2
x
y
−
1
\frac{y^2-x}{2 y^3-2 x y-1}
2
y
3
−
2
x
y
−
1
y
2
−
x
Correct answer
D.
y
2
−
x
2
y
3
−
2
x
y
−
1
\frac{y^2-x}{2 y^3-2 x y-1}
2
y
3
−
2
x
y
−
1
y
2
−
x
Step-by-step solution
We have
y
=
x
+
y
+
x
+
y
+
…
∞
y
=
x
+
y
+
y
\Rightarrowy
2
=
x
+
2
y
⇒
y
2
−
x
=
2
y
⇒
(
y
2
−
x
)
2
=
2
y
⇒
y
4
−
2
x
y
2
+
x
2
=
2
y
⇒
y
4
−
2
x
y
2
−
2
y
+
x
2
=
0
\begin{aligned}& y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+\ldots \infty}}}}\\ & y=\sqrt{x+\sqrt{y+y}} \Rightarrowy^2=x+\sqrt{2 y}\\ \Rightarrow & y²-x=\sqrt{2 y} \Rightarrow\left(y^2-x\right)^2=2 y \\ \Rightarrow & y⁴-2 x y²+x²=2 y \\ \Rightarrow & y⁴-2 x y²-2 y+x²=0 \end{aligned}
⇒
⇒
⇒
y
=
x
+
y
+
x
+
y
+
…
∞
y
=
x
+
y
+
y
\Rightarrowy
2
=
x
+
2
y
y
2
−
x
=
2
y
⇒
(
y
2
−
x
)
2
=
2
y
y
4
−
2
x
y
2
+
x
2
=
2
y
y
4
−
2
x
y
2
−
2
y
+
x
2
=
0
On differentiating w.r.t. to
x
x
x
, we get
4
y
3
d
y
d
x
−
2
y
2
−
4
y
x
d
y
d
x
−
2
d
y
d
x
+
2
x
=
0
d
y
d
x
(
4
y
3
−
4
x
y
−
2
)
=
−
2
x
+
2
y
2
⇒
d
y
d
x
=
y
2
−
x
2
y
3
−
2
x
y
−
1
\begin{aligned}& 4 y³ \frac{d y}{d x}-2 y²-4 y x \frac{d y}{d x}-\frac{2 d y}{d x}+2 x=0 \\ & \frac{d y}{d x}\left(4 y³-4 x y-2\right)=-2 x+2 y² \\ \Rightarrow \quad & \frac{d y}{d x}=\frac{y^2-x}{2 y^3-2 x y-1} \end{aligned}
⇒
4
y
3
d
x
d
y
−
2
y
2
−
4
y
x
d
x
d
y
−
d
x
2
d
y
+
2
x
=
0
d
x
d
y
(
4
y
3
−
4
x
y
−
2
)
=
−
2
x
+
2
y
2
d
x
d
y
=
2
y
3
−
2
x
y
−
1
y
2
−
x
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