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!]#

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Derivatives of powers

%%Q Calculating the derivative of a function is a pretty long-winded 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 short-cuts 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 $f'(x) = nx^{n-1}.$ Equivalently, the derivative of $x^n$ is $nx^{n-1}$.

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

1. \t !r! %%If $f(x) = x^2 \text{ then }f'(x)$ \t ${}= 2x^{2-1}$ \gap[10] \t %%If $f(x) = x^n$ %%then $f'(x) = nx^{n-1} \quad (n = 2)$ \\ \t \t ${}= 2x^1$ \\ \t \t ${}= 2x$. \\ \\ 2. \t !r! %%If $f(x) = x^3 \text{ then } f'(x)$ \t ${}= 3x^{3-1}$ \gap[10] \t %%If $f(x) = x^n$ %%then $f'(x) = nx^{n-1} \quad (n = 3)$ \\ \t \t ${}= 3x^2$. \\ \\ 3. \t !r! %%If $f(x) = x^{-3} \text{ then }f'(x)$ \t ${}= -3x^{-3-1}$ \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 ${}= 1x^{1-1}$ \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 ${}= 0x^{0-1}$ \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.

Some for you

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:

✗

#[WRONG!][¡INCORRECTA!]#

%%Q So what is the correct way to find derivatives like this one?
%%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.

* For purposes of this tutorial, an expression is in power form if no powers of $x$ occur in fractions; for instance

$x^{-1/2} \qquad 4x^{-2} \qquad \dfrac{2}{3}x^{-1} \qquad 3 + x - x^2 \qquad \frac{3}{4}x - 2x^{1/2}$ #[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.)

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.

%%Examples

1. \t $f(x) = \dfrac{1}{x}$ \\ \t Rewrite: \\ \t !r! $f(x)= x^{-1}, \text{ so }f'(x)$ \t ${} = -x^{-1-1}$ \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 ${} = -2x^{-2-1}$ \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 ${} = -0.3x^{-0.3-1}$ \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 ${} = \dfrac{1}{2}x^{1/2-1}$ \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.

Some for you

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 $3x^2$."

The phrase "with respect to $x$" tells us that the independent variable of the function is $x$. We abbreviate the phrase "the derivative with respect to $x$" by the symbol "$d/dx$."

Derivative with respect to x
The notation $\dfrac{d}{dx}$ means the derivative with respect to $x$. Thus, for instance,

$\dfrac{d}{dx}[x^n] = nx^{n-1} \quad$ \t The derivative, with respect to $x$, of $x^n$ equals $nx^{n-1}$. \\ $\dfrac{d}{dx}[x^3] = 3x^2 \quad$ \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

In differential notation, these rules can be written as follows:

In words
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 $\displaystyle \frac{d}{dx}[x^2-x^3+x^5]$ \t ${}= 2x-3x^2+5x^4$ \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 $\displaystyle \frac{d}{dx}[5x^3-4x+7]$ \t ${}= 5(3x^2) - 4(1) + 7(0)$ \gap[10] \t

Combining the properties

\\ \t \t $\displaystyle {}= 15x^2-4$

Some for you

Now try the exercises in Section 4.1 in Applied Calculus or Section 11.1 in Finite Mathematics and Applied Calculus.