Find the derivative of ((1-2x^3)/(1+2x^3))^5

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

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Answer

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

Step by step solution

Problem

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

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

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

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

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

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

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

The derivative of the constant function is equal to zero

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

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

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

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

$5\frac{\left(2x^3+1\right)\left(0-2\cdot 3x^{2}\right)-\left(2\cdot 3x^{\left(3-1\right)}+0\right)\left(1-2x^3\right)}{\left(2x^3+1\right)^2}\left(\frac{1-2x^3}{2x^3+1}\right)^{4}$
7

Subtract the values $3$ and $-1$

$5\frac{\left(2x^3+1\right)\left(0-2\cdot 3x^{2}\right)-\left(2\cdot 3x^{2}+0\right)\left(1-2x^3\right)}{\left(2x^3+1\right)^2}\left(\frac{1-2x^3}{2x^3+1}\right)^{4}$
8

Multiply $3$ times $-2$

$5\frac{\left(2x^3+1\right)\left(0-6x^{2}\right)-\left(6x^{2}+0\right)\left(1-2x^3\right)}{\left(2x^3+1\right)^2}\left(\frac{1-2x^3}{2x^3+1}\right)^{4}$
9

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

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

Multiply $6$ times $-1$

$5\frac{-6\left(2x^3+1\right)x^{2}-6\left(1-2x^3\right)x^{2}}{\left(2x^3+1\right)^2}\left(\frac{1-2x^3}{2x^3+1}\right)^{4}$
11

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$5\frac{-6\left(2x^3+1\right)x^{2}-6\left(1-2x^3\right)x^{2}}{\left(2x^3+1\right)^2}\cdot\frac{\left(1-2x^3\right)^{4}}{\left(2x^3+1\right)^{4}}$
12

Multiplying fractions

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

Answer

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

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.43 seconds

Views:

127