finite mathematics & applied calculus topic summary:

Techniques of differentiation: Part 1 of 2: Sums, products, quotients

Techniques of differentiation: Part 1 of 2: Sums, products, quotients

The

- If $f(x) = x^n$, then $f'(x) = nx^{n-1}$.

- $\dfrac{d}{dx}[x^n] = nx^{n-1}$.

$[f(x) \pm g(x)]' = f'(x) \pm g'(x) \qquad$ | Sum or difference |

$[cf(x)]' = cf'(x) \qquad$ | Constant multiple |

The derivative of c times a function is c times the derivative of the function.

$\displaystyle \frac{d}{dx}[x^{3.2}]$ | ${}=3.2x^{2.2}$ | Power rule | |

$\displaystyle \frac{d}{dx}[4x^{-1}]$ | $\displaystyle {}=4\frac{d}{dx}[x^{-1}]$ | Constant multiple rule | |

${}= 4(-1)x^{-2}$ | Power rule | ||

${}= -4x^{-2}$ | |||

You can get the answer in one step by multiplying the exponent $(-1)$ by the coefficient $(4)$ and then decreasing the exponent by 1. | |||

$\displaystyle \frac{d}{dx}[3x^{1.1}+4x^{-2}]$ | $\displaystyle {}=\frac{d}{dx}[3x^{1.1}]+\frac{d}{dx}[4x^{-2}]$ | Sum rule | |

$\displaystyle {}=3\frac{d}{dx}[x^{1.1}]+4\frac{d}{dx}[x^{-2}]$ | Constant multiple rule | ||

${}= 3(1.1)x^{0.1}+4(-2)x^{-3}$ | Power rule | ||

${}= 3.3x^{0.1}-8x^{-3}$ | |||

You can get the answer in one step by multiplying each exponent by the coefficient and then decreasing the exponent by 1. | |||

$\displaystyle \frac{d}{dx}\left[\frac{1}{x}\right]$ | $\displaystyle {}=\frac{d}{dx}[x^{-1}]$ | Rewrite in power form. | |

$\displaystyle {}=(-1)x^{-2}$ | Power rule | ||

$\displaystyle {}= -\frac{1}{x^2}$ | Rewrite in rational form. | ||

$\displaystyle \frac{d}{dx}\left[\sqrt{x}\right]$ | $\displaystyle {}=\frac{d}{dx}[x^{1/2}]$ | Rewrite in power form. | |

$\displaystyle {}=\frac{1}{2}x^{-1/2}$ | Power rule | ||

$\displaystyle {}= \frac{1}{2\sqrt{x}}$ | Rewrite in rational form. |

Want more practice? Try the online tutorial or the online review exercises.

$\bar{C}$ | $\displaystyle {}=\frac{C(x)}{x}$ |

$\displaystyle {}=\frac{\text{Total cost}}{\text{Number of items}}$ |

- $C(x) = 4x^{0.2}- 0.1x$ €.

- $C'(x) = 0.8x^{-0.8}- 0.1$ € per item.

- $\displaystyle \bar{C}(x) = \frac{C(x)}{x} = \frac{4x^{0.2}- 0.1x}{x} = 4x^{-0.8}- 0.1$ € per item.

If $f$ and $g$ are differentiable functions of $x$, then so is their product $fg$, and

- $\displaystyle \frac{d}{dx}\left[f(x)g(x)\right] = \color{#ff504d}{f'(x)}g(x) + f(x)\color{#ff504d}{g'(x)}.$

If $f$ and $g$ are differentiable functions of $x$, then so is their quotient $f/g$, and

- $\displaystyle \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\color{#ff504d}{f'(x)}g(x) - f(x)\color{#ff504d}{g'(x)}}{[g(x)^2]}.$

- $\displaystyle \frac{d}{dx}\left[x^2(3x-1)\right] = \color{#ff504d}{2x}(3x-1) + x^2(\color{#ff504d}{3}) \qquad$ Derivative of the first × the second, plus the first × derivative of the second

- $\displaystyle \frac{d}{dx}\left[\frac{x^3}{x^2+1}\right] = \frac{\color{#ff504d}{3x^2}(x^2+1) + x^3(\color{#ff504d}{2x})}{(x^2+1)^2} \qquad$

Derivative of the top × the bottom, minus the top × derivative of the bottom | |

bottom squared |

If the CTE says, for instance, that the expression is a sum of two smaller expressions, then apply the rule for sums as a first step. This will leave you having to differentiate the smaller expressions, and you can use the CTE on these, and so on...