Differentiation rules
Differentiate the parts, then combine them according to the structure of the function.
The idea
The limit definition tells us what a derivative means. Differentiation rules let us calculate it efficiently. Choose a rule from how the function is built: a sum, product, or quotient. Compositions will be handled in the next submodule.
Constants, powers, and sums
Rules
Differentiate term by term
For a constant \(c\) and a constant exponent \(n\),
\[\frac{d}{dx}c=0,\qquad \frac{d}{dx}x^n=nx^{n-1}.\]For any real exponent \(n\), the power formula holds for \(x>0\). For positive integer exponents it also holds at zero and at negative inputs; for negative integer exponents it holds at every \(x\ne0\). The case \(n=0\) is the constant rule, with derivative zero.
For rational exponents with an odd denominator in lowest terms, the same power formula also holds at negative inputs. At zero, check whether a finite derivative exists. For other fractional exponents, check the real domain and its endpoint separately. For example, \(\sqrt{x}\) has derivative \(1/(2\sqrt{x})\) for \(x>0\), but no finite right-hand derivative at zero. A formula valid at positive inputs must not be used to assign a derivative at an excluded or singular input.
If \(f\) and \(g\) are differentiable, then
\[\frac{d}{dx}\bigl(cf(x)\bigr)=cf^{\prime}(x),\qquad \frac{d}{dx}\bigl(f(x)\pm g(x)\bigr)=f^{\prime}(x)\pm g^{\prime}(x).\]As in the definition lesson, \(d/dx\) means “differentiate with respect to \(x\).” Multiplying a graph by a constant scales its slopes; adding a constant shifts the graph without changing its slopes.
Example 1
Rewrite before differentiating
For \(x\ne0\), differentiate
\[p(x)=3x^4-\frac{2}{x^2}+7=3x^4-2x^{-2}+7.\]Apply the power rule to each term:
\[p^{\prime}(x)=12x^3+4x^{-3}=12x^3+\frac{4}{x^3}.\]The constant disappears, and the negative exponent produces another negative factor, giving a positive second term.
Products: both factors can change
Rule
Product rule
For differentiable functions \(u\) and \(v\),
\[\frac{d}{dx}(uv)=u^{\prime}v+uv^{\prime}.\]Each term accounts for the change in one factor while using the current value of the other. To see why there are two terms, let a rectangle have side lengths \(u(t)=2+t\) and \(v(t)=3+2t\). At \(t=0\), its area is \(6\). After an increment \(h>0\), two strips and a small corner are added.
Here \(\Delta A=4h+3h+2h^2\), so
\[A^{\prime}(0)=\lim_{h\to0}\frac{(2+h)(3+2h)-6}{h}=\lim_{h\to0}(7+2h)=7.\]This agrees with \(u^{\prime}(0)v(0)+u(0)v^{\prime}(0)=1\cdot3+2\cdot2=7\). The picture shows positive increments; the same algebra works for negative increments near \(0\).
Quotients: order matters
Rule
Quotient rule
Where \(v\ne0\),
\[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime}v-uv^{\prime}}{v^2}.\]Both functions must be differentiable. The numerator uses the derivative of the top factor first; reversing the subtraction changes the sign.
Example 2
Keep the denominator's restriction
For \(q(x)=\frac{x^2+1}{x+1}\), with \(x\ne-1\),
\[q^{\prime}(x)=\frac{(2x)(x+1)-(x^2+1)(1)}{(x+1)^2}=\frac{x^2+2x-1}{(x+1)^2}.\]The square on the denominator belongs to the rule. The original function and this derivative are undefined at \(-1\).
Choose a useful form
Example 3
A product with two changing factors
For \(r(x)=(x^2+2)(x^3-1)\), both factors change with \(x\). The product rule gives
\[r^{\prime}(x)=2x(x^3-1)+(x^2+2)3x^2.\]Expanding first gives \(r(x)=x^5+2x^3-x^2-2\), so term-by-term differentiation gives
\[r^{\prime}(x)=5x^4+6x^2-2x.\]Expanding the product-rule answer produces the same expression. Both methods work; choose the form that makes the calculation clearer.