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Chain Rule
bencejdanko
Given
y
=
f
(
u
)
y = f(u)
y
=
f
(
u
)
and
u
=
g
(
x
)
u = g(x)
u
=
g
(
x
)
, what is the general formula for
d
y
d
x
\frac{dy}{dx}
d
x
d
y
according to the Chain Rule?
d
y
d
x
=
f
′
(
u
)
⋅
g
′
(
x
)
\frac{dy}{dx} = f'(u) \cdot g'(x)
d
x
d
y
=
f
′
(
u
)
⋅
g
′
(
x
)
d
y
d
x
=
f
′
(
u
)
+
g
′
(
x
)
\frac{dy}{dx} = f'(u) + g'(x)
d
x
d
y
=
f
′
(
u
)
+
g
′
(
x
)
d
y
d
x
=
f
′
(
x
)
⋅
g
′
(
u
)
\frac{dy}{dx} = f'(x) \cdot g'(u)
d
x
d
y
=
f
′
(
x
)
⋅
g
′
(
u
)
d
y
d
x
=
f
′
(
g
(
x
)
)
\frac{dy}{dx} = f'(g(x))
d
x
d
y
=
f
′
(
g
(
x
))
Calculate the derivative of
h
(
x
)
=
(
3
x
+
1
)
2
h(x) = (3x + 1)^2
h
(
x
)
=
(
3
x
+
1
)
2
.
6
x
+
1
6x + 1
6
x
+
1
2
(
3
x
+
1
)
2(3x + 1)
2
(
3
x
+
1
)
9
x
2
+
1
9x^2 + 1
9
x
2
+
1
6
(
3
x
+
1
)
6(3x + 1)
6
(
3
x
+
1
)
If
y
=
sin
(
x
2
)
y = \sin(x^2)
y
=
sin
(
x
2
)
, what is
d
y
d
x
\frac{dy}{dx}
d
x
d
y
?
cos
(
2
x
)
\cos(2x)
cos
(
2
x
)
2
x
cos
(
x
2
)
2x \cos(x^2)
2
x
cos
(
x
2
)
2
x
sin
(
x
2
)
2x \sin(x^2)
2
x
sin
(
x
2
)
cos
(
x
2
)
\cos(x^2)
cos
(
x
2
)
Find the derivative of
f
(
x
)
=
e
5
x
f(x) = e^{5x}
f
(
x
)
=
e
5
x
.
e
5
x
e^{5x}
e
5
x
1
5
e
5
x
\frac{1}{5}e^{5x}
5
1
e
5
x
5
x
e
5
x
−
1
5xe^{5x-1}
5
x
e
5
x
−
1
5
e
5
x
5e^{5x}
5
e
5
x
Differentiate
y
=
ln
(
cos
x
)
y = \ln(\cos x)
y
=
ln
(
cos
x
)
.
−
tan
x
-\tan x
−
tan
x
1
sin
x
\frac{1}{\sin x}
s
i
n
x
1
1
cos
x
\frac{1}{\cos x}
c
o
s
x
1
−
sin
x
-\sin x
−
sin
x
If
f
(
x
)
=
1
+
x
2
f(x) = \sqrt{1 + x^2}
f
(
x
)
=
1
+
x
2
, what is
f
′
(
x
)
f'(x)
f
′
(
x
)
?
x
1
+
x
2
\frac{x}{\sqrt{1 + x^2}}
1
+
x
2
x
1
2
1
+
x
2
\frac{1}{2\sqrt{1 + x^2}}
2
1
+
x
2
1
x
1
+
x
2
x\sqrt{1+x^2}
x
1
+
x
2
2
x
1
+
x
2
\frac{2x}{\sqrt{1 + x^2}}
1
+
x
2
2
x
In the function
h
(
x
)
=
sin
(
cos
(
x
)
)
h(x) = \sin(\cos(x))
h
(
x
)
=
sin
(
cos
(
x
))
, which is the "inner" function?
x
x
x
sin
(
x
)
\sin(x)
sin
(
x
)
cos
(
x
)
\cos(x)
cos
(
x
)
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