Derive the function 16/(x^2+4) with respect to x

\frac{d}{dx}\left(\frac{16}{x^2+4}\right)

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Answer

$\frac{-32x}{\left(4+x^2\right)^2}$

Step by step solution

Problem

$\frac{d}{dx}\left(\frac{16}{x^2+4}\right)$
1

Applying the quotient rule 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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\left(4+x^2\right)\frac{d}{dx}\left(16\right)-16\frac{d}{dx}\left(4+x^2\right)}{\left(4+x^2\right)^2}$
2

The derivative of the constant function is equal to zero

$\frac{0\left(4+x^2\right)-16\frac{d}{dx}\left(4+x^2\right)}{\left(4+x^2\right)^2}$
3

Any expression multiplied by $0$ is equal to $0$

$\frac{0-16\frac{d}{dx}\left(4+x^2\right)}{\left(4+x^2\right)^2}$
4

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

$\frac{0-16\left(\frac{d}{dx}\left(4\right)+\frac{d}{dx}\left(x^2\right)\right)}{\left(4+x^2\right)^2}$
5

The derivative of the constant function is equal to zero

$\frac{0-16\left(0+\frac{d}{dx}\left(x^2\right)\right)}{\left(4+x^2\right)^2}$
6

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

$\frac{0-16\left(0+2x\right)}{\left(4+x^2\right)^2}$
7

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

$\frac{-32x}{\left(4+x^2\right)^2}$

Answer

$\frac{-32x}{\left(4+x^2\right)^2}$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.21 seconds

Views:

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