Tutorial: Derivatives of powers, sums, and constant multiples
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(This topic is also in Section 4.1 in Applied Calculus or Section 11.1 in Finite Mathematics and Applied Calculus) #[I don't like this new tutorial. Take me back to the old tutorial!][No me gusta este nueve tutorial. ¡Regresame al tutorial más viejo!]#

Derivatives of powers
%%Q Calculating the derivative of a function is a pretty longwinded process. Isn't there an easier method?%%A For practically all the functions you have seen, the short answer is "yes". In this section we study shortcuts that will allow you to write down the derivative of powers of $x$ (including fractional and negative powers) as well as sums and constant multiples of powers of $x$, such as polynomials. We start with the rule that gives the derivative of a power of $x$:
Power rule
If $f(x) = x^n$, where $n$ is any constant, then $f'(x) = nx^{n1}.$ Equivalently, the derivative of $x^n$ is $nx^{n1}$. Power rule in words
The derivative of $x$ raised to a constant power equals that power times $x$ raised to the power minus 1. Proof of the power rule
Want to see a proof of the power rule? #[][]# Click here.
If $f(x) = x^n$, where $n$ is any constant, then $f'(x) = nx^{n1}.$ Equivalently, the derivative of $x^n$ is $nx^{n1}$. Power rule in words
The derivative of $x$ raised to a constant power equals that power times $x$ raised to the power minus 1. Proof of the power rule
Want to see a proof of the power rule? #[][]# Click here.
%%Examples
Some for you
1. \t !r! %%If $f(x) = x^2 \text{ then }f'(x)$ \t ${}= 2x^{21}$ \gap[10] \t %%If $f(x) = x^n$ %%then $f'(x) = nx^{n1} \quad (n = 2)$
\\ \t \t ${}= 2x^1$
\\ \t \t ${}= 2x$.
\\
\\ 2. \t !r! %%If $f(x) = x^3 \text{ then } f'(x)$ \t ${}= 3x^{31}$ \gap[10] \t %%If $f(x) = x^n$ %%then $f'(x) = nx^{n1} \quad (n = 3)$
\\ \t \t ${}= 3x^2$.
\\
\\ 3. \t !r! %%If $f(x) = x^{3} \text{ then }f'(x)$ \t ${}= 3x^{31}$ \gap[10] \t Works for negative exponents as well. $(n = 3)$.
\\ \t \t ${}= 3x^{4}$.
\\
\\ 4. \t !r! %%If $f(x) = x \text{ then }f'(x)$ \t ${}= 1x^{11}$ \gap[10] \t $x = x^1$, so $n = 1$.
\\ \t \t ${}= x^0$
\\ \t \t ${}= 1$.
\\
\\ 5. \t !r! %%If $f(x) = 1 \text{ then }f'(x)$ \t ${}= 0x^{01}$ \gap[10] \t $1 = x^01$, so $n = 0$
\\ \t \t ${}= 0$.
\\
Here is a table showing these results. We suggest you make a copy of the table; we will be adding to it as we go.
Powers of x in the denominator
Warning The power rule does not apply to powers of $x$ in the denominator; for instance the following claim is wrong:

%%If $\displaystyle f(x) = \frac{1}{x^3}$, %%then $\displaystyle f'(x) = \frac{1}{3x^2}\qquad$
✗ #[WRONG!][¡INCORRECTA!]#
%%A The trick is to write the given expression in power form* before taking its derivative:So, for instance, we calculate the derivative correctly as follows:
$\displaystyle f(x) = \frac{1}{x^3}$ \t $\displaystyle {} = x^{3}\quad$ \t First convert to power form.
%%Therefore,
\\ $\displaystyle f'(x)$ \t $\displaystyle {}= 3x^{4}\quad$ \t Now use the power rule (the power of $x$ is in the numerator so the power rule applies).
\\ \t $\displaystyle {} = \frac{3}{x^{4}}\quad$ \t Convert back to the original positive exponent form if desired.
Dealing with powers of x in the denominator
As we said above, the power rule does not apply to powers of $x$ in the denominator. To find the derivative of $f(x) = \dfrac{1}{x^n}$, rewrite $f(x)$ as $x^{n}$ and use the power rule, which does apply to powers of $x$ in the numerator.
As we said above, the power rule does not apply to powers of $x$ in the denominator. To find the derivative of $f(x) = \dfrac{1}{x^n}$, rewrite $f(x)$ as $x^{n}$ and use the power rule, which does apply to powers of $x$ in the numerator.
%%Examples
Some for you
1. \t $f(x) = \dfrac{1}{x}$
\\ \t Rewrite:
\\ \t !r! $f(x)= x^{1}, \text{ so }f'(x)$ \t ${} = x^{11}$ \gap[10] \t Power rule
\\ \t \t ${}= x^{2}$
\\ \t \t ${}= \dfrac{1}{x^2}$ \t Convert back to positive exponent form if desired
\\
\\ 2. \t $f(x) = \dfrac{1}{x^2}$
\\ \t Rewrite:
\\ \t !r! $f(x)= x^{2}, \text{ so }f'(x)$ \t ${} = 2x^{21}$ \gap[10] \t Power rule
\\ \t \t ${}= 2x^{3}$
\\ \t \t ${}= \dfrac{2}{x^3}$ \t Convert back to positive exponent form if desired
\\ 3. \t $f(x) = \dfrac{1}{x^{0.3}}$
\\ \t Rewrite:
\\ \t !r! $f(x)= x^{0.3}, \text{ so }f'(x)$ \t ${} = 0.3x^{0.31}$ \gap[10] \t Power rule
\\ \t \t ${}= 0.3x^{1.3}$
\\ \t \t ${}= \dfrac{0.3}{x^{1.3}}$ \t Convert back to positive exponent form if desired
\\
\\ 4. \t $f(x) = \dfrac{1}{x^{1/2}}$
\\ \t Rewrite:
\\ \t !r! $f(x)= x^{1/2}, \text{ so }f'(x)$ \t ${} = \dfrac{1}{2}x^{1/21}$ \gap[10] \t Power rule
\\ \t \t ${}= \dfrac{1}{2}x^{1/2}$
\\ \t \t ${}= \dfrac{1}{2x^{1/2}}$ \t Convert to positive exponent form if desired
\\
Here is the above table extended to show these results. Again, we suggest you make a copy of the table adding to it as we go.
Differential notation
Differential notation is based on an abbreviation for the phrase "the derivative with respect to $x$." For example, we learned above that if $f(x) = x^3$, then $f\prime(x) = 3x^2$. When we say "$f\prime(x) = 3x^2$," we mean:

"The derivative of $x^3$ with respect to $x$ equals $3x^2$."
Derivative with respect to x
The notation $\dfrac{d}{dx}$ means the derivative with respect to $x$. Thus, for instance,
The notation $\dfrac{d}{dx}$ means the derivative with respect to $x$. Thus, for instance,
$\dfrac{d}{dx}[x^n] = nx^{n1} \quad$ \t The derivative, with respect to $x$, of $x^n$ equals $nx^{n1}$.
\\ $\dfrac{d}{dx}[x^3] = 3x^2 \quad$ \t The derivative, with respect to $x$, of $x^3$ equals $3x^2$.
\\ $\dfrac{d}{dx}[1] = 0 \quad$ \t The derivative, with respect to $x$, of $1$ equals $0$.
\\ $\dfrac{d}{dx}\left[\dfrac{1}{x}\right] = \dfrac{1}{x^2} \quad$ \t The derivative, with respect to $x$, of $\dfrac{1}{x}$ equals $\dfrac{1}{x^2}$.
Derivatives of sums, differences, and constant multiples
#[We can find the derivative of more complicated expressions using the following: ][A continuación, podemos determinar las derivadas de funciones más complicadas aplicando las siguientes reglas:]#
Rule for sums, differences, and constant multiples
If $f\prime(x)$ and $g\prime(x)$ exist, and $c$ is a constant, then
The derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the derivatives.
In other words, to find the derivative of a sum (or difference) of several functions, just find the derivative of each function, and add (or subtract) the answers. The derivative of $c$ times a function is $c$ times the derivative of the function.
In other words, to find the derivative of a constant times a function, just find the derivative of the function, and multiply by the constant.
If $f\prime(x)$ and $g\prime(x)$ exist, and $c$ is a constant, then

a. $\displaystyle [f(x) \pm g(x)]\prime = f'(x) \pm g\prime(x)$
b. $\displaystyle [c\,f(x)]\prime = c\,f\prime(x) $

a. $\displaystyle \frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)]$
b. $\displaystyle \frac{d}{dx}[c\,f(x)] = c\frac{d}{dx}[f(x)] $
The derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the derivatives.
In other words, to find the derivative of a sum (or difference) of several functions, just find the derivative of each function, and add (or subtract) the answers. The derivative of $c$ times a function is $c$ times the derivative of the function.
In other words, to find the derivative of a constant times a function, just find the derivative of the function, and multiply by the constant.
%%Examples
1. \t $\displaystyle \frac{d}{dx}[1+x^3]$ \t ${}= 0 + 3x^2$ \gap[10] \t
Property a.
\\ \t \t ${}= 3x^2$
\\
\\ 2. \t $\displaystyle \frac{d}{dx}[x^2x^3+x^5]$ \t ${}= 2x3x^2+5x^4$ \gap[10] \t
Property a also works for three or more terms
\\
3. \t $\displaystyle \frac{d}{dx}[4x^3]$ \t $\displaystyle {}= 4\frac{d}{dx}[x^3]$ \gap[10] \t
Property b (take out the constant).
\\ \t \t ${}= 4(3x^2)$
\\ \t \t ${}= 12x^2$ \t The effect is to multiply the exponent (3) by the coefficient (4).
\\
4. \t $\displaystyle \frac{d}{dx}[12]$ \t $\displaystyle {}= \frac{d}{dx}[12 \cdot 1]$ \gap[10] \t
Rewrite 12 as 12 × 1.
\\ \t \t $\displaystyle {}= 12\frac{d}{dx}[1]$ \gap[10] \t Property b (take out the constant).
\\ \t \t ${}= 12(0)$ \t The derivative of 1 is 0.
\\ \t \t ${}= 0$ \t In general, the derivative of any constant is zero.
\\
5. \t $\displaystyle \frac{d}{dx}\left[\frac{4}{x}\right]$ \t $\displaystyle {}= \frac{d}{dx}\left[4 \cdot \frac{1}{x}\right]$ \gap[10] \t
Rewrite the quotient as a product.
\\ \t \t $\displaystyle {}= 4\frac{d}{dx}\left[\frac{1}{x}\right]$ \t Property b (take out the constant).
\\ \t \t $\displaystyle {}= 4\left[\frac{1}{x^2}\right]$ \t See the table of derivatives above.
\\ \t \t $\displaystyle {}= \frac{4}{x^2}$
\\
\\ 6. \t $\displaystyle \frac{d}{dx}[5x^34x+7]$ \t ${}= 5(3x^2)  4(1) + 7(0)$ \gap[10] \t
Some for you
Combining the properties
\\ \t \t $\displaystyle {}= 15x^24$
Now try the exercises in Section 4.1 in Applied Calculus or Section 11.1 in Finite Mathematics and Applied Calculus.
Copyright © 2018 Stefan Waner and Steven R. Costenoble
$x^{1/2} \qquad 4x^{2} \qquad \dfrac{2}{3}x^{1} \qquad 3 + x  x^2 \qquad \frac{3}{4}x  2x^{1/2}$
#[are all in power form, but not][son todos en la forma potencia, pero no]#$\dfrac{3x}{4} \qquad \dfrac{1}{x^3} \qquad \dfrac{2}{4x^{3}} \qquad x + \dfrac{1}{x} \qquad \dfrac{2}{3x^{1}}$
(See the %%powerformtut for details and practice.)