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Joint Entrance Examination
Mathematics
Limits, Continuity and Differentiability
Question 276
Question 276 of 401
Medium
MCQ
If
x
=
e
y
+
e
y
+
⋯
to
∞
x=e^{y+e^{y+\cdots\text{ to }\infty}}
x
=
e
y
+
e
y
+
⋯
to
∞
,
x
>
0
x>0
x
>
0
, then
d
y
d
x
\dfrac{dy}{dx}
d
x
d
y
is
A
1
+
x
x
\dfrac{1+x}{x}
x
1
+
x
B
1
x
\dfrac1x
x
1
C
1
−
x
x
\dfrac{1-x}{x}
x
1
−
x
D
x
1
+
x
\dfrac{x}{1+x}
1
+
x
x
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Solutions · 1
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Official solution
x
=
e
y
+
x
⇒
ln
x
=
y
+
x
x=e^{y+x}\Rightarrow\ln x=y+x
x
=
e
y
+
x
⇒
ln
x
=
y
+
x
. Differentiating:
1
x
=
d
y
d
x
+
1
⇒
d
y
d
x
=
1
x
−
1
=
1
−
x
x
\dfrac1x=\dfrac{dy}{dx}+1\Rightarrow\dfrac{dy}{dx}=\dfrac1x-1=\dfrac{1-x}{x}
x
1
=
d
x
d
y
+
1
⇒
d
x
d
y
=
x
1
−
1
=
x
1
−
x
.
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