Find the derivative using logarithmic differentiation method $\frac{d}{dx}\left(\ln\left(\sqrt{\frac{4x+1}{x^2}}\right)^{\tan\left(2x\right)}\right)$
Related Videos
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
◻2
◻◻
√◻
√
◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc
asin
acos
atan
acot
asec
acsc
sinh
cosh
tanh
coth
sech
csch
asinh
acosh
atanh
acoth
asech
acsch
Share this solution with your friends!
Function Plot
Plotting: $\left(\sec\left(2x\right)^2\ln\left(\left(\frac{1}{2}\ln\left(4x+1\right)-\ln\left(x\right)\right)^2\right)+\left(\frac{2}{4x+1}+\frac{-1}{x}\right)\frac{\tan\left(2x\right)}{\frac{1}{2}\ln\left(4x+1\right)-\ln\left(x\right)}\right)\left(\frac{1}{2}\ln\left(4x+1\right)-\ln\left(x\right)\right)^{\tan\left(2x\right)}$
SnapXam
A2
Answer Assistant
beta
Got a different answer? Verify it!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
◻2
◻◻
√◻
√
◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc
asin
acos
atan
acot
asec
acsc
sinh
cosh
tanh
coth
sech
csch
asinh
acosh
atanh
acoth
asech
acsch
How to improve your answer: