MAC 2311
Test Two Proofs.
1) Prove:
d
⎡sin x ⎤⎦ = cos x
dx ⎣
(
) ( )
(
)
f x+h − f x
sin x + h − sin x
d
sin x cosh+ cos x sinh− sin x
⎡⎣sin x ⎤⎦ = lim
= lim
= lim
=
h→0
h→0
h→0
dx
h
h
h
⎡
⎛ cosh− 1⎞
⎛ sinh ⎞ ⎤
⎛
⎛
cosh− 1⎞
sinh ⎞
lim ⎢sin x ⎜
+
cos
x
=
sin
x
lim
+
cos
x
lim
⎥
⎜⎝ h ⎟⎠
⎜⎝ h→0 h ⎟⎠
⎜⎝ h→0 h ⎟⎠ =
h→0
h ⎟⎠
⎝
⎣
⎦
(
(sin x )(0) + (cos x )(1) = cos x
2) Prove:
)
(
Therefore,
)
d
⎡sin x ⎤⎦ = cos x
dx ⎣
d
⎡cos x ⎤⎦ = − sin x
dx ⎣
(
) ( )
(
)
f x+h − f x
cos x + h − cos x
d
cos x cosh− sin x sinh− cos x
⎡⎣cos x ⎤⎦ = lim
= lim
= lim
=
h→0
h→0
h→0
dx
h
h
h
⎡
⎛ cosh− 1⎞
⎛ sinh ⎞ ⎤
⎛
⎛
cosh− 1⎞
sinh ⎞
lim ⎢cos x ⎜
−
sin
x
=
cos
x
lim
−
sin
x
lim
⎥
⎟⎠
⎜⎝ h ⎟⎠
⎜⎝ h→0 h ⎟⎠
⎜⎝ h→0 h ⎟⎠ =
h→0
h
⎝
⎣
⎦
(
(cos x )(0) − (sin x )(1) = − sin x
3) Prove:
)
Therefore,
(
)
d
⎡cos x ⎤⎦ = − sin x
dx ⎣
d
⎡⎣ tanx ⎤⎦ = sec 2 x
dx
(
)(
) (
)(
) (
)(
) (
)(
)
)(
) (
)(
) (
)(
) (
)(
)
′
′
cos x cos x − sin x − sin x
d
d ⎡ sin x ⎤ cos x sin x − sin x cos x
⎡⎣ tan x ⎤⎦ =
=
=
=
⎢
⎥
2
dx
dx ⎣ cos x ⎦
cos x
cos2 x
cos2 x + sin2 x
1
d
⎡⎣ tan x ⎤⎦ = sec 2 x
=
= sec 2 x
Therefore,
2
2
dx
cos x
cos x
4) Prove:
d
⎡cot x ⎤⎦ = − csc 2 x
dx ⎣
(
′
′
sin x − sin x − cos x cos x
d
d ⎡ cos x ⎤ sin x cos x − cos x sin x
⎡⎣cot x ⎤⎦ =
=
=
=
⎢
⎥
2
dx
dx ⎣ sin x ⎦
sin x
sin2 x
− sin2 x − cos2 x
sin2 x + cos2 x
1
d
⎡⎣cot x ⎤⎦ = − csc 2 x
=
−
= − 2 = − csc 2 x
Therefore,
2
2
dx
sin x
sin x
sin x
5) Prove:
d
⎡sec x ⎤⎦ = sec x tanx
dx ⎣
(
)( ) ( )(
) (
)( ) ( )(
)
′
′
cos x 0 − 1 − sinx
d
d ⎡ 1 ⎤ cos x 1 − 1 cos x
sinx
⎡⎣sec x ⎤⎦ =
=
=
=
=
⎢
⎥
2
2
dx
dx ⎣ cos x ⎦
cos x
cos x
cos2 x
1
sinx
d
⎡sec x ⎤⎦ = sec x tanx
⋅
= sec x tanx
Therefore,
cos x cos x
dx ⎣
6) Prove:
d
⎡csc x ⎤⎦ = − csc x cot x
dx ⎣
(
)( ) ( )(
) (
)( ) ( )(
)
′
′
sinx 0 − 1 cos x
d
d ⎡ 1 ⎤ sinx 1 − 1 sinx
cos x
⎡⎣csc x ⎤⎦ =
=
=
=− 2 =
⎢
⎥
2
2
dx
dx ⎣ sinx ⎦
sin x
sin x
sin x
1 cos x
d
⎡csc x ⎤⎦ = − csc x cot x
−
⋅
= − csc x cot x
Therefore,
sinx sinx
dx ⎣