derivative Log[absolute[Sec[x] Tan[x]]]
In[15]:=
Assuming úLogø is the natural logarithm È Use the base 10 logarithm instead
Derivative:
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Possible derivation:
â
HlogH secHxL tanHxL¤LL
âx
â
HlogH secHxL tanHxL¤LL ‡
Use the chain rule,
âx
âu
âx
u
, where u ‡ tanHxL secHxL¤ and True:
â
H secHxL tanHxL¤L
âx
secHxL tanHxL¤
â
H tanHxL secHxL¤L ‡
Use the chain rule,
âx
¢
Abs HtanHxL secHxLL I
âu
âx
¢
Abs HuL, where u ‡ tanHxL secHxL and True:
â
HtanHxL secHxLLM
âx
secHxL tanHxL¤
â
Hu vL ‡ v
âu
âv
, where u ‡ secHxL and v ‡ tanHxL:
âx
âx
âx
â
â
¢
Abs HtanHxL secHxLL IsecHxL I HtanHxLLM + tanHxL I HsecHxLLMM
âx
âx
Use the product rule,
+u
secHxL tanHxL¤
The derivative of tanHxL is sec2 HxL:
¢
Abs HtanHxL secHxLL ItanHxL I
â
HsecHxLLM + sec2 HxL secHxLM
âx
secHxL tanHxL¤
The derivative of secHxL is tanHxL secHxL:
Isec3 HxL+tanHxL HtanHxL secHxLLM Abs¢ HtanHxL secHxLL
secHxL tanHxL¤
secHxL is the secant function »
z¤ is the absolute value of z »
logHxL is the natural logarithm »
Plots:
20
2
Untitled-1
20
10
x
-2 Π
-Π
2Π
Π
-10
-20
min
max
20
10
x
-20
10
-10
20
-10
-20
min
max
Alternate forms:
More
2 tanHxL + cotHxL
secHxL HsinHxL + cscHxLL
Isin2 HxL + 1M cscHxL secHxL
cotHxL is the cotangent function »
Alternate form assuming x is real:
sinH2 xL
4 sinHxL cosHxL
-
cosH2 xL + 1
HcosH2 xL - 1L HcosH2 xL + 1L
Roots:
Step-by-step solution
x ‡ 2 Π n - ä tanh-1
3-2
2
» 2 H3.14159 n - 0.440687 äL , n Î Z
x ‡ 2 Π n + ä tanh-1
3-2
2
» 2 H3.14159 n + 0.440687 äL , n Î Z
x ‡ 2 Π n - ä tanh-1
3+2
2
» 2 H3.14159 n - H1.5708 + 0.440687 äLL , n Î Z
x ‡ 2 Π n + ä tanh-1
3+2
2
» 2 H3.14159 n + H1.5708 + 0.440687 äLL , n Î Z
tanh-1 HxL is the inverse hyperbolic tangent function »
Z is the set of integers »
Properties as a real function:
Approximate forms
Domain:
Π
8x Î R : Hx + Π > 2 Π c2 and 2 x + Π < 4 Π c2 and c2 , c1 Î ZL or K-
2
< x - 2 Π c2 < 0 and c2 , c1 Î ZO
Untitled-1
Π
or K0 < x - 2 Π c2 <
2
Π
and c2 , c1 Î ZO or K
2
< x - 2 Π c2 < Π and c2 , c1 Î ZO<
Periodicity:
Periodic in x with period Π
Series expansion at x=0:
1
x
29 x3
5x
+
+
3
50 x5
+
45
189
+ OIx6 M
Indefinite integral:
à HcscHxL secHxL + tanHxLL â x ‡ logHsinHxLL - 2 logHcosHxLL + constant
Step-by-step solution
3