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# Find the derivative using the product rule $\frac{d}{dx}\left(2x+1\right)$

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asinh
acosh
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acoth
asech
acsch

##  Final answer to the problem

$2$
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##  Step-by-step Solution 

How should I solve this problem?

• Find the derivative using the product rule
• Find the derivative using the definition
• Find the derivative using the quotient rule
• Find the derivative using logarithmic differentiation
• Find the derivative
• Integrate by partial fractions
• Product of Binomials with Common Term
• FOIL Method
• Integrate by substitution
• Integrate by parts
Can't find a method? Tell us so we can add it.
1

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(1\right)$

Learn how to solve product rule of differentiation problems step by step online.

$\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(1\right)$

Learn how to solve product rule of differentiation problems step by step online. Find the derivative using the product rule d/dx(2x+1). The derivative of a sum of two or more functions is the sum of the derivatives of each function. Apply the product rule for differentiation: (f\cdot g)'=f'\cdot g+f\cdot g', where f=. The derivative of the constant function (2) is equal to zero. The derivative of the constant function (1) is equal to zero.

##  Final answer to the problem

$2$

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

SnapXam A2

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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
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

###  Main Topic: Product Rule of differentiation

The product rule is a formula used to find the derivatives of products of two or more functions. It may be stated as $(f\cdot g)'=f'\cdot g+f\cdot g'$

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