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# Find the derivative $\frac{d}{dx}\left(\frac{1}{1+3x^2}\right)$

## Step-by-step Solution

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### Videos

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

Problem to solve:

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

Specify the solving method

1

Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by $h(x) = \frac{f(x)}{g(x)}$, where ${g(x) \neq 0}$, then $h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}$

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

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

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

Learn how to solve quotient rule of differentiation problems step by step online. Find the derivative (d/dx)(1/(1+3x^2)). Apply the quotient rule for differentiation, which states that if f(x) and g(x) are functions and h(x) is the function defined by {\displaystyle h(x) = \frac{f(x)}{g(x)}}, where {g(x) \neq 0}, then {\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}. The derivative of the constant function (1) is equal to zero. The derivative of a sum of two or more functions is the sum of the derivatives of each function. The derivative of the constant function (1) is equal to zero.

$\frac{-6x}{\left(1+3x^2\right)^2}$
SnapXam A2

### beta Got another answer? Verify it!

Go!
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

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