How do you find the derivative of 1/(1+x^4)?

Nov 16, 2016

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

Explanation:

Given function "f(x)" (hypothetical) $\Rightarrow \frac{1}{1 + {x}^{4}} \Leftarrow$ which is a fraction/rational;

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)$ should be calculated via application of the Quotient Rule of differentiation.

Quotient Rule

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)$ when $f \left(x\right) = \frac{a}{b}$ is differentiated into $\frac{\frac{d}{\mathrm{dx}} \left(a\right) \setminus \times b - a \setminus \times \frac{d}{\mathrm{dx}} \left(b\right)}{b} ^ 2$

Application

$\frac{\frac{d}{\mathrm{dx}} \left(1\right) \setminus \times \left(1 + {x}^{4}\right) - \frac{d}{\mathrm{dx}} \left(\frac{1}{1 + {x}^{4}}\right) = 1 \setminus \times \frac{d}{\mathrm{dx}} \left(1 + {x}^{4}\right)}{1 + {x}^{4}} ^ 2$

$\setminus \rightarrow \frac{\left(0\right) \left(1 + {x}^{4}\right) - 1 \setminus \times \left(0 + 4 {x}^{3}\right)}{1 + {x}^{4}} ^ 2$

$\setminus \rightarrow \frac{0 - 1 \setminus \times \left(4 {x}^{3}\right)}{1 + {x}^{4}} ^ 2$

$\setminus \rightarrow \frac{- 4 {x}^{3}}{1 + {x}^{4}} ^ 2$