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Find the implicit derivative $\frac{d}{dx}\left(y=x\arctan\left(2x\right)\right)$

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Solving: $\frac{d}{dx}\left(y=x\arctan\left(2x\right)\right)$
Solve instead: $\frac{dy}{dx}\left(y=x\tan^{-1}\left(2x\right)\right)$

Solution

Formulas

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Final answer to the problem

$y^{\prime}=\arctan\left(2x\right)+\frac{2x}{1+4x^2}$
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Learn how to solve inverse trigonometric functions differentiation problems step by step online. Find the implicit derivative d/dx(y=xarctan(2x)). Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable. The derivative of the linear function is equal to 1. Apply the product rule for differentiation: (f\cdot g)'=f'\cdot g+f\cdot g', where f=x and g=\arctan\left(2x\right). The derivative of the linear function is equal to 1.

Final answer to the problem

$y^{\prime}=\arctan\left(2x\right)+\frac{2x}{1+4x^2}$

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Function Plot

Plotting: $y^{\prime}=\arctan\left(2x\right)+\frac{2x}{1+4x^2}$

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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:

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Main Topic: Inverse trigonometric functions differentiation

Describes how to find the derivative of the inverse of the trigonometric functions sine, cosine, tangent, etc.

Used Formulas

See formulas (4)

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