Derivative rules, with the reasoning

The derivative measures output change per unit of input change. The rules below follow what happens when we add, multiply, or compose functions. Assume the functions being differentiated are differentiable at the point in question.

Sum: changes add

If $f$ changes by approximately $f'(x)h$ and g by $g'(x)h$, their sum changes by approximately $(f'(x)+g'(x))h$. Thus

$$(f+g)'=f'+g',\qquad (cf)'=cf'$$

when c is constant. For $f(x)=3x^2+5x-7$, differentiate each term: $f'(x)=6x+5$. The constant −7 shifts every output equally, so it contributes no change.

Power: expand the change

For a square,

$$(x+h)^2-x^2=2xh+h^2.$$

Divide by h to get $2x+h$. The extra h approaches zero, leaving $2x$. When x and h are positive, a growing square provides a picture of this algebra:

The two strips contribute 2xh. The corner contributes h². After division by h, the corner contributes h, which approaches zero.

For a positive integer n, expand n copies of $(x+h)$. The terms containing exactly one h add up to $nx^{n-1}h$. Terms containing two or more factors of h vanish after division by h and taking the limit:

$$\frac{d}{dx}x^n=nx^{n-1}.$$

That argument proves the integer case. The rule also extends to real powers on $x>0$; particular powers have larger domains. For example, $(1/x)'=-1/x^2$ for $x\ne0$, while $(\sqrt{x})'=1/(2\sqrt{x})$ for $x>0$.

Product: both factors can change

Let u and v depend on x. Write their actual changes as $\Delta u$ and $\Delta v$. Expanding gives the exact identity

$$(u+\Delta u)(v+\Delta v)-uv =v\Delta u+u\Delta v+\Delta u\Delta v.$$

Divide by the input change h. The first two terms approach $vu'$ and $uv'$. The last approaches zero: it contains a product of two changes, each proportional to h to first order. Therefore

$$(uv)'=u'v+uv'.$$

For $x^2x^3$, this gives $2x\cdot x^3+x^2\cdot3x^2=5x^4$. This agrees with differentiating $x^5$ directly. Multiplying the two derivatives would incorrectly give $6x^3$.

Chain: follow the intermediate input

Consider $y=(3x+1)^2$. Introduce $u=3x+1$, so the calculation is $x\to u\to y$.

At $x=1$, we have $u=4$. A small change h in x produces approximately 3h in u. Near $u=4$, squaring changes the output by approximately $2u=8$ times its input change. Combining the two gives $8\cdot3h=24h$.

This is the chain rule:

$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} =2(3x+1)\cdot3.$$

Evaluate the outer derivative at the intermediate value u, then multiply by the inner derivative. At $x=1$, the result is 24. Expanding first gives $9x^2+6x+1$, whose derivative $18x+6$ confirms it.

Try it: differentiate (2x − 1)³ at x = 2

The intermediate value is $u=2(2)-1=3$. The outer derivative is $3u^2=27$ and the inner derivative is 2. The answer is $27\cdot2=54$. Returning 27 would omit the effect of x on u.

Its own slope

Exponentials and logarithms

For $a>0$, the average rate of $a^x$ factors as

$$\frac{a^{x+h}-a^x}{h}=a^x\frac{a^h-1}{h}.$$

The second factor does not depend on x. Its limit is $\ln a$, so $(a^x)'=(\ln a)a^x$. The special base $e\approx2.718$ has $\ln e=1$, giving $(e^x)'=e^x$.

For $x>0$, differentiate $e^{\ln x}=x$ using the chain rule. This gives $e^{\ln x}(\ln x)'=1$, hence $(\ln x)'=1/x$.

An exponent that is the variable, as in $2^x$, requires a different rule from a constant exponent, as in $x^2$.

The quotient rule

When $v(x)\ne0$, apply the product rule to $u(1/v)$ and the chain rule to $1/v$:

$$\left(\frac uv\right)'=\frac{u'v-uv'}{v^2}.$$

The condition $v(x)\ne0$ matters because the original quotient must be defined.

Next

Use derivatives to investigate maxima and minima, or practice choosing a rule.