Tutorial: Derivatives of powers, sums, and constant multiples
(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 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.
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
Some for you
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.
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:
-
$\displaystyle \frac{d}{dx}\left[\frac{1}{x^3}\right] = \frac{1}{3x^2}\qquad$
✗ #[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:
$\displaystyle \frac{d}{dx}\left[\frac{1}{x^3}\right]$ \t $\displaystyle {} = \frac{d}{dx}\left[x^{-3}\right]\quad$ \t First convert to power form.
\\ \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 ${} = -x^{-1-1}$ \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.
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,
$\dfrac{d}{dx}[x^n] = nx^{n-1} \quad$ \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
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 $\displaystyle \frac{d}{dx}[x^2-x^3+x^5]$ \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
Some for you
Combining the properties
\\ \t \t $\displaystyle {}= 15x^2-4$
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.)