Differentiate a power series term by term
The slope of a polynomial that copies a curve
The chalk curve in the upper graph is \(f(x)=\dfrac1{1+x}\). It equals the series \(1-x+x^2-x^3+\cdots\) for \(x\) between \(-1\) and 1, and each piece of the series is a term.
The orange curve is the polynomial that keeps only the first terms. The slider sets its degree, which is the highest power of \(x\) that the polynomial keeps. At a higher degree, the orange curve is closer to the chalk curve.
The lower graph shows slopes. The chalk curve is the slope of \(f\). The orange curve is the slope of the polynomial, which you find by differentiating each term.
At \(x=0.5\), the slope of \(f\) is \(-\tfrac49\), about \(-0.444\). There, the slope of the polynomial is \(-0.5625\) at degree 5 and about \(-0.448\) at degree 11.
The slope of the polynomial moves toward the slope of \(f\) as you keep more terms. This means that differentiating each term of the series gives a series for the slope of \(f\).
Any power series
A power series is a sum of terms \(c_nx^n\), where each number \(c_n\) is a coefficient. It converges for every \(x\) closer to 0 than a distance \(R\), which is the radius of convergence.
The teal bar on the x-axis shows these values of \(x\). Inside the bar, both orange curves move toward the chalk curves when you raise the degree.
Press the first button to use a series with a smaller radius. The teal bar narrows, and both orange curves leave the chalk curves at its ends, because differentiating each term does not change the radius.
Notation
Write the series with a sum sign. The sum adds the terms for \(n=0,1,2,\dots\)
\[\begin{aligned} f(x)&=\sum_{n=0}^{\infty}c_nx^n \\ &=c_0+c_1x+c_2x^2+\cdots \end{aligned}\]To differentiate one term, use the power rule. The exponent \(n\) becomes a factor, and the new exponent is \(n-1\).
\[\dfrac{d}{dx}\left(c_nx^n\right)=\cy{nc_n}\,x^{n-1}\]The term \(c_0\) is a constant, so its derivative is 0, and the new sum starts at \(n=1\).
\[\begin{aligned} f'(x)&=\sum_{n=1}^{\infty}nc_nx^{n-1} \\ &=c_1+2c_2x+3c_3x^2+\cdots \end{aligned}\]For a series centered at \(a\), the same rule holds with powers of \(x-a\).
For every \(x\) with \(|x|\lt R\), the derivative of \(f\) equals the sum of the derivatives of its terms. The new series has the same radius \(R\).
When it fails
Move \(x\) beyond 1 and keep more terms. The orange curves move farther from the chalk curves.
The series of \(f\) converges only for \(|x|<1\). Outside this interval, the series has no sum, so differentiating its terms tells you nothing about \(f\).
Example 1: the sine series
The series of \(\sin x\) is \(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\), and it converges for every \(x\). Differentiate it term by term.
Differentiate each term with the power rule.
\[\begin{aligned} &\dfrac{d}{dx}\left(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}\right) \\ &\quad=1-\dfrac{3x^2}{3!}+\dfrac{5x^4}{5!} \end{aligned}\]Simplify each coefficient. Because \(3!=3\cdot2!\), the fraction \(\frac3{3!}\) equals \(\frac1{2!}\). In the same way, \(\frac5{5!}=\frac1{4!}\).
\[=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots\]This is the series of \(\cos x\), the derivative of \(\sin x\). The radius did not change, so it converges for every \(x\).
Example 2: the exponential series
The series of \(e^{2x}\) is \(1+2x+2x^2+\frac43x^3+\cdots\). Find the series of \(\dfrac{d}{dx}e^{2x}\).
The dashed orange line in the upper graph is the tangent of \(e^{2x}\) at \(x=0\), and its slope is 2. The series of the derivative must equal 2 at \(x=0\).
\[\cy{2}\]A tempting wrong move: Multiply each coefficient by its exponent. Leave the exponents as they are.
\[0+2x+4x^2+4x^3+\cdots\]Every term of this series is 0 at \(x=0\), so the red curve in the lower graph starts at 0. The slope of \(e^{2x}\) at \(x=0\) is 2, not 0.
Multiply each coefficient by its exponent. Then lower each exponent by 1. The constant term 1 has the derivative 0, so it leaves the sum.
\[2+4x+4x^2+\tfrac83x^3+\cdots\]Every coefficient is twice the matching coefficient of the original series, so the new series is twice the series of \(e^{2x}\), which is \(2e^{2x}\).
\[\begin{aligned} &2\left(1+2x+2x^2+\tfrac43x^3+\cdots\right) \\ &\quad=2e^{2x} \end{aligned}\]
Practice
The series of \(\dfrac1{1-x^2}\) is \(1+x^2+x^4+x^6+\cdots\) for \(|x|<1\). Find the first four nonzero terms of the series of its derivative. Then write the derivative as a fraction.
In the lower graph, the chalk curve is the slope of the function. It is 0 at \(x=0\), so the series of the derivative has no constant term.
Solution
Differentiate each term. The constant term 1 leaves the sum, the term \(x^2\) gives \(2x\), and the term \(x^4\) gives \(4x^3\).
\[2x+4x^3+6x^5+8x^7+\cdots\]To write the fraction, differentiate \((1-x^2)^{-1}\) with the chain rule.
\[\begin{aligned} &\dfrac{d}{dx}(1-x^2)^{-1} \\ &\quad=-(1-x^2)^{-2}\cdot(-2x) \\ &\quad=\dfrac{2x}{(1-x^2)^2} \end{aligned}\]The original series converges when \(|x^2|<1\), that is \(|x|<1\). The new series has the same radius.
Questions
The graph shows \(g(x)=\dfrac1{1-3x}\) and its tangent at \(x=0\), the dashed orange line. The series of \(g\) is \(1+3x+9x^2+27x^3+\cdots\). The dashed vertical line marks where \(g\) is undefined.
What is the constant term in the series of \(g'(x)\)?
- 3
- 1
- 0
Answer
3
What is the radius of convergence of the series of \(g'(x)\)?
- \(\dfrac13\)
- 1
- 3
Answer
\(\dfrac13\)
A power series has the coefficients \(c_n=\dfrac{(-1)^n}{n+1}\). What is the coefficient of \(x^4\) in the series of its derivative?
- \(-\dfrac56\)
- \(-\dfrac16\)
- \(\dfrac45\)
- \(\dfrac15\)
Answer
\(-\dfrac56\)
Summary
To differentiate a power series, you multiply each coefficient by its exponent and lower each exponent by 1, so the constant term becomes 0.
Inside the interval of convergence, the new series converges to the derivative, and its radius is the same.
You use it to write a series for the slope of a function that you know as a series.
This is one of five skills in topic 10.15 of Which & Why: AP Calculus.