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Question 0821
Maclaurin Series
2021 Paper 1 Question 7 Variant
Question
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It is given that
y
=
e
3
tan
x
,
{y=\mathrm{e}^{3 \tan x},}
y
=
e
3
t
a
n
x
,
for
x
∈
R
.
{x \in \mathbb{R}.}
x
∈
R
.
(a)
Show that
d
2
y
d
x
2
=
y
−
1
(
d
y
d
x
)
2
+
2
3
ln
y
d
y
d
x
.
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=y^{-1}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 +\frac{2}{3} \ln y \frac{\mathrm{d}y}{\mathrm{d}x}.
d
x
2
d
2
y
=
y
−
1
(
d
x
d
y
)
2
+
3
2
ln
y
d
x
d
y
.
[4]
(b)
Find the first 4 terms of the Maclaurin expansion of
e
3
tan
x
.
{\mathrm{e}^{3 \tan x}.}
e
3
t
a
n
x
.
[5]
Answer
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