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# Find the derivative using the product rule $\frac{d}{dx}\left(\left(2x+1\right)^5\left(x^4-3\right)^6\right)$

## Step-by-step Solution

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### Videos

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24x^{3}\left(2x+1\right)^5\left(x^4-3\right)^{5}$
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## Step-by-step Solution

Problem to solve:

$\frac{d}{dx}\left(2x+1\right)^5\left(x^4-3\right)^6$

Choose the solving method

1

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=\left(2x+1\right)^5$ and $g=\left(x^4-3\right)^6$

$\frac{d}{dx}\left(\left(2x+1\right)^5\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$5\left(2x+1\right)^{\left(5-1\right)}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$

Subtract the values $5$ and $-1$

$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
2

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$5\left(2x+1\right)^{\left(5-1\right)}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$

Subtract the values $5$ and $-1$

$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$

$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{\left(6-1\right)}\frac{d}{dx}\left(x^4-3\right)$

Subtract the values $6$ and $-1$

$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
3

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
4

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

$5\left(2x+1\right)^{4}\left(\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(1\right)\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$

The derivative of the constant function ($1$) is equal to zero

$5\left(2x+1\right)^{4}\left(\frac{d}{dx}\left(2x\right)+0\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$

$x+0=x$, where $x$ is any expression

$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
5

The derivative of the constant function ($1$) is equal to zero

$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$

The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function

$10\left(2x+1\right)^{4}\frac{d}{dx}\left(x\right)\left(x^4-3\right)^6$

The derivative of the linear function is equal to $1$

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6$
6

The derivative of the linear function times a constant, is equal to the constant

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
7

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

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\left(\frac{d}{dx}\left(x^4\right)+\frac{d}{dx}\left(-3\right)\right)$

The derivative of the constant function ($1$) is equal to zero

$5\left(2x+1\right)^{4}\left(\frac{d}{dx}\left(2x\right)+0\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$

$x+0=x$, where $x$ is any expression

$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\left(\frac{d}{dx}\left(x^4\right)+0\right)$

$x+0=x$, where $x$ is any expression

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
8

The derivative of the constant function ($-3$) is equal to zero

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\cdot 4x^{\left(4-1\right)}\left(2x+1\right)^5\left(x^4-3\right)^{5}$

Subtract the values $4$ and $-1$

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\cdot 4x^{3}\left(2x+1\right)^5\left(x^4-3\right)^{5}$

Multiply $6$ times $4$

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24x^{3}\left(2x+1\right)^5\left(x^4-3\right)^{5}$
9

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24x^{3}\left(2x+1\right)^5\left(x^4-3\right)^{5}$

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24x^{3}\left(2x+1\right)^5\left(x^4-3\right)^{5}$
SnapXam A2

Go!
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7
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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
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

### Tips on how to improve your answer:

$\frac{d}{dx}\left(2x+1\right)^5\left(x^4-3\right)^6$

### Main topic:

Product Rule of differentiation

~ 0.1 s