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Science
:
Math Exam Resources/Courses/MATH100/December 2011/Question 01 (i)/Solution 1
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From UBC Wiki
<
Science:Math Exam Resources
|
Courses/MATH100
|
December 2011
|
Question 01 (i)
Using the fact that
d
d
x
sin
−
1
(
x
)
=
1
1
−
x
2
,
{\displaystyle {\frac {d}{dx}}\sin ^{-1}(x)={\frac {1}{\sqrt {1-x^{2}}}},}
and the chain rule, we get
d
d
x
y
=
d
d
x
sin
−
1
(
3
x
+
1
)
=
1
1
−
(
3
x
+
1
)
2
⋅
d
d
x
(
3
x
+
1
)
=
3
1
−
(
3
x
+
1
)
2
.
{\displaystyle {\begin{aligned}{\frac {d}{dx}}y&={\frac {d}{dx}}\sin ^{-1}(3x+1)\\&={\frac {1}{\sqrt {1-(3x+1)^{2}}}}\cdot {\frac {d}{dx}}(3x+1)\\&={\frac {3}{\sqrt {1-(3x+1)^{2}}}}.\end{aligned}}}