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Joint Entrance Examination
Mathematics
Limits, Continuity and Differentiability
Question 258
Question 258 of 401
Medium
MCQ
If
f
′
(
x
)
=
sin
(
log
x
)
f'(x)=\sin(\log x)
f
′
(
x
)
=
sin
(
lo
g
x
)
and
y
=
f
(
2
x
+
3
3
−
2
x
)
y=f\left(\dfrac{2x+3}{3-2x}\right)
y
=
f
(
3
−
2
x
2
x
+
3
)
, then
d
y
d
x
\dfrac{dy}{dx}
d
x
d
y
equals
A
sin
(
log
2
x
+
3
3
−
2
x
)
\sin\left(\log\dfrac{2x+3}{3-2x}\right)
sin
(
lo
g
3
−
2
x
2
x
+
3
)
B
12
(
3
−
2
x
)
2
\dfrac{12}{(3-2x)^2}
(
3
−
2
x
)
2
12
C
12
(
3
−
2
x
)
2
sin
(
log
2
x
+
3
3
−
2
x
)
\dfrac{12}{(3-2x)^2}\sin\left(\log\dfrac{2x+3}{3-2x}\right)
(
3
−
2
x
)
2
12
sin
(
lo
g
3
−
2
x
2
x
+
3
)
D
12
(
3
−
2
x
)
2
cos
(
log
2
x
+
3
3
−
2
x
)
\dfrac{12}{(3-2x)^2}\cos\left(\log\dfrac{2x+3}{3-2x}\right)
(
3
−
2
x
)
2
12
cos
(
lo
g
3
−
2
x
2
x
+
3
)
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Official solution
\[\frac{dy}{dx}=f'\left(\frac{2x+3}{3-2x}\right)\cdot\frac{d}{dx}\left(\frac{2x+3}{3-2x}\right)=\sin\left(\log\frac{2x+3}{3-2x}\right)\cdot\frac{12}{(3-2x)^2}\]
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