Why derivative patterns help
A Taylor polynomial of degree \(n\) requires the values of a function and its first \(n\) derivatives at a chosen center. When \(n\) is large, differentiating repeatedly and independently is slow and makes sign and factorial errors more likely. Many familiar functions have a pattern that lets us generate all the required derivative values in an organized way.
The goal here is not to find a new type of approximation. It is to compute the coefficients of a Taylor polynomial more efficiently by identifying what happens to the derivatives as the order increases.
Write \(h=x-a\) when it makes the expressions easier to read. The coefficient of \(h^k\) is \(f^{(k)}(a)/k!\), so an efficient calculation has two parts: find a rule for \(f^{(k)}(a)\), then divide each value by its matching factorial. A pattern in the derivatives does not remove the factorial in the Taylor formula.
A general way to organize the calculation
Before calculating, identify what changes when one more derivative is taken. The result might repeat in a cycle, as with sine and cosine; it might gain a constant factor, as with an exponential; or it might gain a new factor in a product, as with a real power. Once that pattern is clear, evaluate it at the center and fill the coefficients in order.
Look for a cycle, a constant multiplier, or a recurrence that relates one derivative to the next.
Use the pattern to determine \(f^{(k)}(a)\) for each order needed.
Use \(k!\) for the coefficient of \((x-a)^k\), including \(0!=1\).
Differentiate the resulting polynomial and confirm that its derivatives at \(a\) match the function’s derivatives through order \(n\).
Exponential derivatives: the same factor at every step
For \(f(x)=e^{cx}\), where \(c\) is a constant, each differentiation multiplies the current expression by \(c\). In fact, for every nonnegative integer \(k\),
This follows by induction. It holds for \(k=0\), since \(f^{(0)}(x)=e^{cx}\) and \(c^0=1\). If it holds for \(k\), differentiating gives \(f^{(k+1)}(x)=c^k c e^{cx}=c^{k+1}e^{cx}\), establishing the next case. The same pattern applies to \(e^{c(x-a)}\); its \(k\)th derivative is \(c^k e^{c(x-a)}\).
Worked Example: A degree-five Maclaurin polynomial for \(e^{2x}\)
Find the Taylor polynomial of degree five centered at zero. Here \(f(x)=e^{2x}\), so \(f^{(k)}(x)=2^k e^{2x}\). At zero, \(e^0=1\), and therefore the derivative values for orders zero through five are \(1,2,4,8,16,32\).
Divide each value by the factorial for its order: \(0!=1\), \(1!=1\), \(2!=2\), \(3!=6\), \(4!=24\), and \(5!=120\). The coefficients are \(1,2,2,4/3,2/3,4/15\), respectively. Thus
Check the coefficients by differentiating at zero: \(P_5(0)=1\), \(P_5'(0)=2\), \(P_5''(0)=2(2)=4\), \(P_5'''(0)=6(4/3)=8\), \(P_5^{(4)}(0)=24(2/3)=16\), and \(P_5^{(5)}(0)=120(4/15)=32\). These match \(f^{(k)}(0)=2^k\) at every order from zero through five.
Trigonometric derivatives: keep track of the cycle
The derivatives of sine and cosine repeat every four differentiations. Starting from \(\sin x\), the cycle is \(\sin x,\cos x,-\sin x,-\cos x\), after which the first expression returns. Starting from \(\cos x\), it is \(\cos x,-\sin x,-\cos x,\sin x\). This follows by differentiating each expression once and recording the result. At a nonzero center, the cycle still applies; evaluate the appropriate sine or cosine value at that center.
Worked Example: A degree-six Taylor polynomial for sine at \(\pi/6\)
Let \(f(x)=\sin x\) and take \(a=\pi/6\). The derivative cycle gives \(f,f',f'',f''',f^{(4)},f^{(5)},f^{(6)}=\sin x,\cos x,-\sin x,-\cos x,\sin x,\cos x,-\sin x\). Since \(\sin(\pi/6)=1/2\) and \(\cos(\pi/6)=\sqrt{3}/2\), the derivative values at the center, in order, are \(1/2,\sqrt{3}/2,-1/2,-\sqrt{3}/2,1/2,\sqrt{3}/2,-1/2\).
Divide those values by \(0!=1,1!=1,2!=2,3!=6,4!=24,5!=120,6!=720\), respectively. The resulting coefficients are \(1/2,\sqrt{3}/2,-1/4,-\sqrt{3}/12,1/48,\sqrt{3}/240,-1/1440\). Therefore, with \(h=x-\pi/6\),
For a coefficient \(c_k\) of \(h^k\), the \(k\)th derivative of that term at \(h=0\) is \(k!c_k\). Here the values \(k!c_k\), from \(k=0\) to \(6\), are exactly \(1/2,\sqrt{3}/2,-1/2,-\sqrt{3}/2,1/2,\sqrt{3}/2,-1/2\). Thus the polynomial’s derivatives at \(x=\pi/6\) agree with the sine derivative cycle through degree six.
Logarithmic derivatives: alternate signs and factorials
For \(f(x)=\ln x\), the domain is \(x>0\). The first derivative is \(1/x\). Further differentiations give \(-1/x^2\), \(2/x^3\), and \(-6/x^4\). For every integer \(k\geq 1\), the pattern is
To see why the formula continues, suppose it holds at order \(k\). Differentiating multiplies by \(-k\) and changes \(x^{-k}\) to \(x^{-k-1}\). The new coefficient is \((-1)^{k-1}(k-1)!(-k)=(-1)^k k!\), as required for order \(k+1\). At a center \(a>0\), the Taylor coefficient for \(k\geq1\) is consequently \((-1)^{k-1}(k-1)!/(k!a^k)=(-1)^{k-1}/(k a^k)\).
Worked Example: A degree-five Taylor polynomial for \(\ln x\) at \(1\)
At \(a=1\), \(f(1)=\ln 1=0\). The derivative formula gives \(f'(1)=1\), \(f''(1)=-1\), \(f'''(1)=2\), \(f^{(4)}(1)=-6\), and \(f^{(5)}(1)=24\). Dividing in turn by \(1!,2!,3!,4!,5!\) gives coefficients \(1,-1/2,1/3,-1/4,1/5\). Therefore
To check, the derivatives of this polynomial at \(1\) are \(0,1,-1,2,-6,24\) at orders zero through five: each coefficient of \((x-1)^k\), multiplied by \(k!\), gives the corresponding value. These agree with \(\ln x\) and its derivatives at \(1\).
Real powers: build a falling product
For a real exponent \(\alpha\), consider \(f(x)=(1+x)^\alpha\) on an interval where \(1+x>0\). Repeated differentiation gives
Each derivative lowers the exponent by one and multiplies by the exponent currently in front. In particular, at \(x=0\), the derivative value is the product \(\alpha(\alpha-1)\cdots(\alpha-k+1)\). Dividing by \(k!\) gives the Taylor coefficient. This method avoids expanding the product afresh through a long chain of differentiation.
Worked Example: A degree-five Taylor polynomial for \(\sqrt{1+x}\)
Set \(\alpha=1/2\). The derivative values at zero, beginning with \(f(0)=1\), are obtained from the falling products:
Now divide by the matching factorials: \(1/2\) by \(1!\) gives \(1/2\); \(-1/4\) by \(2!\) gives \(-1/8\); \(3/8\) by \(3!\) gives \(1/16\); \(-15/16\) by \(4!\) gives \(-5/128\); and \(105/32\) by \(5!\) gives \(7/256\). Thus
Check by multiplying each coefficient by its factorial: the resulting derivative values at zero are \(1,1/2,-1/4,3/8,-15/16,105/32\), exactly the values obtained from the falling products. The square root function is defined here for \(x>-1\), which includes the center \(0\).
Common pitfall: confusing derivatives with coefficients
A frequent error is to use the derivative values themselves as the Taylor coefficients. For example, for \(\ln x\) at \(1\), the third derivative is \(f'''(1)=2\). Writing \(2(x-1)^3\) as the cubic term is incorrect: the Taylor coefficient is \(f'''(1)/3!=2/6=1/3\). The correct cubic term is \((x-1)^3/3\). Differentiating that term three times gives \(6/3=2\) at the center, as required.
Another risk is losing track of a cycle. With sine, the second derivative is \(-\sin x\), not \(\sin x\); the minus sign changes every other pair of entries. Write the cycle explicitly before substituting the center rather than trying to recall several signs at once.
Check Your Understanding
Use the derivative patterns and coefficient checks from this tutorial to answer the following questions.
- For \(f(x)=e^{3x}\), what is \(f^{(4)}(0)\), and what is the coefficient of \(x^4\) in its degree-four Maclaurin polynomial?
- List the first four derivatives of \(\cos x\), beginning with the function itself. What is the fourth derivative?
- For \(\ln x\) centered at \(a>0\), give the coefficient of \((x-a)^k\) for \(k\geq1\).
- Write the falling product that gives the sixth derivative of \((1+x)^\alpha\) at zero.
- Why is it incorrect to use \(f^{(k)}(a)\) directly as the coefficient of \((x-a)^k\)?