Find the derivative $\frac{d}{dx}\left(\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}\right)$

Step-by-step Solution

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Final answer to the problem

$\left(\frac{2}{x}-2+\frac{-2\mathrm{cosh}\left(2x\right)}{\mathrm{sinh}\left(2x\right)}\right)\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}$
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Apply the formula: $\frac{d}{dx}\left(x\right)$$=y=x$, where $d/dx=\frac{d}{dx}$, $d/dx?x=\frac{d}{dx}\left(\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}\right)$ and $x=\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}$

$y=\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}$

Learn how to solve les limites de l'infini problems step by step online.

$y=\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}$

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Learn how to solve les limites de l'infini problems step by step online. Find the derivative d/dx((4x^2)/(e^(2x)sinh(2x))). Apply the formula: \frac{d}{dx}\left(x\right)=y=x, where d/dx=\frac{d}{dx}, d/dx?x=\frac{d}{dx}\left(\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}\right) and x=\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}. Apply the formula: y=x\to \ln\left(y\right)=\ln\left(x\right), where x=\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}. Apply the formula: y=x\to y=x, where x=\ln\left(\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}\right) and y=\ln\left(y\right). Apply the formula: \ln\left(y\right)=x\to \frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(x\right), where x=\ln\left(4x^2\right)-2x-\ln\left(\mathrm{sinh}\left(2x\right)\right).

Final answer to the problem

$\left(\frac{2}{x}-2+\frac{-2\mathrm{cosh}\left(2x\right)}{\mathrm{sinh}\left(2x\right)}\right)\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}$

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Function Plot

Plotting: $\left(\frac{2}{x}-2+\frac{-2\mathrm{cosh}\left(2x\right)}{\mathrm{sinh}\left(2x\right)}\right)\frac{4x^2}{e^{2x}\mathrm{sinh}\left(2x\right)}$

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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
Dx
|◻|
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

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