Derivative Rules and Techniques

Derivative rules match expression structure

Each derivative rule applies to a particular operation used to build a function.

Read the outermost operation first. The expression $(x^2+1)(x^3-2)$ is a product; $(3x-1)^5$ is a composition; $(x^2+1)/(x-1)$ is a quotient. Similar-looking symbols can therefore need different rules.

The rules assume the component derivatives exist at the points where they are used; differentiability must be checked separately. Domain restrictions survive simplification. A derivative formula cannot create a value at an input excluded from the original function.

Constant, sum and power rules

Core derivative rules

Core derivative rules
Function	Derivative	Conditions
$c$	$0$	Constant $c$ .
$x^n$	$nx^{n-1}$	For real $n$ , use $x>0$ . Integer powers also apply on their differentiability domains.
$cf(x)$	$cf'(x)$	Constant $c$ ; $f$ differentiable.
$f(x)+g(x)$	$f'(x)+g'(x)$	Both functions differentiable.

For example, the derivative of $x^7$ is $7x^6$ .

For a positive integer power, the rule applies to every real input. Negative integer powers exclude zero. For an arbitrary real power, a convenient common setting is $x>0$ ; fractional powers may have additional real-domain extensions whose endpoints need their own checks.

\frac{d}{dx}\left(\frac1x\right)=-\frac1{x^2}\quad(x\ne0),\qquad\frac{d}{dx}\sqrt{x}=\frac1{2\sqrt{x}}\quad(x>0)

Constants use the constant rule, including the case of a function equal to 1. There is no need to interpret an undefined intermediate expression such as $0\,x^{-1}$ at zero.

Product rule

If $f$ and $g$ are differentiable, both factors contribute to the change in their product:

\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)

For this example, hold one factor unchanged in each term:

\frac{d}{dx}[(x^2+1)(x^3-2)]=2x(x^3-2)+(x^2+1)3x^2

=5x^4+3x^2-4x

Expanding the product first gives $x^5+x^3-2x^2-2$ , whose derivative is the same. This is a useful independent check. Multiplying the two derivatives alone would give $6x^3$ and miss both required contributions.

Quotient rule

For differentiable $f,g$ with $g(x)\ne0$ ,

\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}

In the original fraction, the numerator derivative is $2x$ and the denominator derivative is 1:

\frac{d}{dx}\frac{x^2+1}{x-1}=\frac{2x(x-1)-(x^2+1)}{(x-1)^2}

=\frac{x^2-2x-1}{(x-1)^2}\quad(x\ne1)

Subtract the entire second product in the numerator; parentheses protect its signs. Dividing the separate derivatives, $2x/1$ , is not a quotient rule.

Chain rule

A composition changes in two stages: the input changes the inner function, and that inner output changes the outer function. If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , the chain rule gives

\frac{d}{dx}f(g(x))=f'(g(x))\,g'(x)

Keep the inner expression inside the outer derivative

For $\sin(x^2)$ , the outer derivative is cosine and the inner derivative is $2x$ :

\frac{d}{dx}\sin(x^2)=\cos(x^2)\cdot2x

For $(3x-1)^5$ , first differentiate the fifth power, retaining its argument, then multiply by the derivative of $3x-1$ :

\frac{d}{dx}(3x-1)^5=5(3x-1)^4\cdot3=15(3x-1)^4

The inner derivative multiplies the result. It does not replace the argument of the outer derivative.

Exponential, logarithmic and trigonometric rules

Selected common derivatives

Selected derivatives with their domains
Function	Derivative	Condition
$e^x$	$e^x$	All real $x$ .
$a^x$	$a^x\ln a$	$a>0$ constant.
$\ln x$	$1/x$	$x>0$ .
$\sin x$	$\cos x$	Radians.
$\cos x$	$-\sin x$	Radians.
$\tan x$	$\sec^2x$	Radians; $\cos x\ne0$ .

With composition, apply the chain rule as well. Thus $\frac{d}{dx}e^{2x}=2e^{2x}$ and $\frac{d}{dx}\ln(x^2+1)=\frac{2x}{x^2+1}$ .

The trigonometric derivative formulas in the table use radian inputs. A degree-valued expression such as $\sin(\pi x/180)$ has derivative $(\pi/180)\cos(\pi x/180)$ with respect to the numerical degree variable $x$ .

The logarithm $\ln x$ requires $x>0$ . For $\ln(x^2+1)$ the argument is positive for every real input, so its derivative is defined everywhere. If $a>0$ is constant, $d(a^x)/dx=a^x\ln a$ ; the special base $e$ has $\ln e=1$ .

Implicit differentiation

Some relations do not conveniently solve for y first. Differentiate both sides with respect to x while treating y as a function of x. From $x^2+y^2=25$ :

2x+2y\frac{dy}{dx}=0

So $\frac{dy}{dx}=-\frac{x}{y}$ where y≠0. At points with y=0 the graph has vertical tangents instead of finite slopes.

The circle is not globally the graph of one function $y=f(x)$ , but its upper and lower arcs can be treated as local functions where $y\ne0$ . At $(3,4)$ the slope is $-3/4$ , so the tangent is $y-4=-\tfrac34(x-3)$ .

The factor $dy/dx$ appears because $y^2$ is a composition with $y(x)$ . Treating $y$ as a constant while differentiating with respect to $x$ would discard the dependence that defines the curve.

Higher derivatives

Differentiating again gives the second derivative, written $f”(x)$ . It measures how the first derivative changes and supports acceleration and concavity analysis. For $f(x)=x^4$ , $f”(x)=12x^2$ .

f(x)=x^4,\quad f'(x)=4x^3,\quad f”(x)=12x^2

A second derivative is not the square of the first derivative. Here $[f'(x)]^2=16x^6$ , which is a different function. When the input carries a unit, each further differentiation divides by that input unit again. Squaring a derivative is a different operation.

Practice questions

Differentiate $(3x-1)^5$ .
Differentiate $\sin(x^2)$ with a radian argument.
For $x^2+y^2=25$ , find $dy/dx$ where $y\ne0$ .

Check your answers

The outer derivative is $5(3x-1)^4$ and the inner derivative is 3. Their product is $15(3x-1)^4$ .
The outer derivative contributes $\cos(x^2)$ and the inner derivative contributes $2x$ , giving $2x\cos(x^2)$ .
Treat $y$ as a function of $x$ : $2x+2y(dy/dx)=0$ . Hence $dy/dx=-x/y$ where $y\ne0$ .

Sources and further reading

These references support the definitions, theorem assumptions and standard conventions used in this article.

OpenStax Calculus Volume 1: Differentiation Rules. Basic derivative rules.
OpenStax Calculus Volume 1: The Chain Rule. Differentiating compositions.
OpenStax Calculus Volume 1: Implicit Differentiation. Implicit relations and dy/dx.
Wolfram Language: D. Symbolic differentiation reference.