Skip to main content
Textbookish
Home
Textbooks
Sign In
Textbooks
/
Calculus
/
Question 10.3
Question 10.3
Solutions
TZ
leumasic
Official
3 months ago
We have
1
)
(
tan
x
)
′
=
(
sin
x
cos
x
)
′
=
cos
2
x
+
sin
2
x
cos
2
x
=
1
cos
2
x
=
sec
2
x
.
\begin{align*} 1)\quad (\tan x)'&=\left(\frac{\sin x}{\cos x}\right)' \\ &=\frac{\cos^2x+\sin^2x}{\cos^2x} \\ &=\frac{1}{\cos^2x} \\ &=\sec^2x. \end{align*}
1
)
(
tan
x
)
′
=
(
cos
x
sin
x
)
′
=
cos
2
x
cos
2
x
+
sin
2
x
=
cos
2
x
1
=
sec
2
x
.
2
)
(
cot
x
)
′
=
(
cos
x
sin
x
)
′
=
−
sin
2
x
−
cos
2
x
sin
2
x
=
−
csc
2
x
.
\begin{align*} 2)\quad (\cot x)'&=\left(\frac{\cos x}{\sin x}\right)' \\ &=\frac{-\sin^2x-\cos^2x}{\sin^2x} \\ &=-\csc^2x. \end{align*}
2
)
(
cot
x
)
′
=
(
sin
x
cos
x
)
′
=
sin
2
x
−
sin
2
x
−
cos
2
x
=
−
csc
2
x
.
3
)
(
sec
x
)
′
=
(
1
cos
x
)
′
=
sin
x
cos
2
x
=
sec
x
tan
x
.
\begin{align*} 3)\quad (\sec x)'&=\left(\frac{1}{\cos x}\right)' \\ &=\frac{\sin x}{\cos^2x} \\ &=\sec x\tan x. \end{align*}
3
)
(
sec
x
)
′
=
(
cos
x
1
)
′
=
cos
2
x
sin
x
=
sec
x
tan
x
.
4
)
(
csc
x
)
′
=
(
1
sin
x
)
′
=
−
cos
x
sin
2
x
=
−
csc
x
cot
x
.
\begin{align*} 4)\quad (\csc x)'&=\left(\frac{1}{\sin x}\right)' \\ &=\frac{-\cos x}{\sin^2x} \\ &=-\csc x\cot x. \end{align*}
4
)
(
csc
x
)
′
=
(
sin
x
1
)
′
=
sin
2
x
−
cos
x
=
−
csc
x
cot
x
.
0
Copy link
Submit a solution
Optional • Markdown
Sign in to share your solution for this question.
Sign in
Navigate
Q 10.3
Prev
Next
Navigate
Q 10.3
Prev
Next
