Skip to main content

CLP-1 Differential Calculus

Section 2.6 Using the Arithmetic of Derivatives – Examples

Subsection 2.6.1 Using the Arithmetic of Derivatives – Examples

In this section we illustrate the computation of derivatives using the arithmetic of derivatives — Theorems 2.4.22.4.3 and 2.4.5. To make it clear which rules we are using during the examples we will note which theorem we are using:
LIN to stand for “linearity” ddx{αf(x)+βg(x)}=αf(x)+βg(x) Theorem 2.4.2
PR to stand for “product rule” ddx{f(x)g(x)}=f(x)g(x)+f(x)g(x) Theorem 2.4.3
QR to stand for “quotient rule” ddx{f(x)g(x)}=f(x)g(x)f(x)g(x)g(x)2 Theorem 2.4.5
We’ll start with a really easy example.

Example 2.6.1. ddx{4x+7}.

ddx{4x+7}=4ddx{x}+7ddx{1}LIN=41+70=4
where we have used LIN with f(x)=x, g(x)=1, α=4, β=7.

Example 2.6.2. ddx{x(4x+7)}.

Continuing on from the previous example, we can use the product rule and the previous result to compute
ddx{x(4x+7)}=xddx{4x+7}+(4x+7)ddx{x}PR=x4+(4x+7)1=8x+7
where we have used the product rule PR with f(x)=x and g(x)=4x+7.

Example 2.6.3. ddx{x4x+7}.

In the same vein as the previous example, we can use the quotient rule to compute
ddx{x4x+7}=(4x+7)ddx{x}xddx{4x+7}(4x+7)2QR=(4x+7)1x4(4x+7)2=7(4x+7)2
where we have used the quotient rule QR with f(x)=x and g(x)=4x+7.
Now for a messier example.

Example 2.6.4. Some examples should be messy.

Differentiate
f(x)=x2x+13x+1
This problem looks nasty. But it isn’t so hard if we just build it up a bit at a time.
  • First, f(x) is the ratio of
    f1(x)=xandf2(x)=2x+13x+1
    If we can find the derivatives of f1(x) and f2(x), we will be able to get the derivative of f(x) just by applying the quotient rule. The derivative, f1(x)=1, of f1(x) is easy, so let’s work on f2(x).
  • The function f2(x) is the linear combination
    f2(x)=2f3(x)+f4(x)withf3(x)=xandf4(x)=13x+1
    If we can find the derivatives of f3(x) and f4(x), we will be able to get the derivative of f2(x) just by applying linearity (Theorem 2.4.2). The derivative, f3(x)=1, of f3(x) is easy. So let’s work of f4(x).
  • The function f4(x) is the ratio
    f4(x)=1f5(x)withf5(x)=3x+1
    If we can find the derivative of f5(x), we will be able to get the derivative of f4(x) just by applying the special case the quotient rule (Corollary 2.4.6). The derivative of f5(x) is easy.
  • So we have completed breaking down f(x) into easy pieces. It is now just a matter of reversing the break down steps, putting everything back together, starting with the easy pieces and working up to f(x). Here goes.
    f5(x)=3x+1 so ddxf5(x)=3ddxx+ddx1=31+0=3LINf4(x)=1f5(x) so ddxf4(x)=f5(x)f5(x)2=3(3x+1)2QRf2(x)=2f3(x)+f4(x) so ddxf2(x)=2f3(x)+f4(x)=23(3x+1)2LINf(x)=f1(x)f2(x) so ddxf(x)=f1(x)f2(x)f1(x)f2(x)f2(x)2QR=1[2x+13x+1]x[23(3x+1)2][2x+13x+1]2
    Oof!
  • We now have an answer. But we really should clean it up, not only to make it easier to read, but also because invariably such computations are just small steps inside much larger computations. Any future computations involving this expression will be a lot easier and less error prone if we clean it up now. Cancelling the 2x and the 2x in
    1[2x+13x+1]x[23(3x+1)2]=2x+13x+12x+3x(3x+1)2=13x+1+3x(3x+1)2
    and multiplying both the numerator and denominator by (3x+1)2 gives
    f(x)=13x+1+3x(3x+1)2[2x+13x+1]2 (3x+1)2(3x+1)2=(3x+1)+3x[2x(3x+1)+1]2=6x+1[6x2+2x+1]2.
While the linearity theorem (Theorem 2.4.2) is stated for a linear combination of two functions, it is not difficult to extend it to linear combinations of three or more functions as the following example shows.

Example 2.6.5. Linearity of the derivative of three or more functions.

We’ll start by generalising linearity to three functions.
ddx{aF(x)+bG(x)+cH(x)}=ddx{ a[F(x)] + 1[bG(x)+cH(x)] }=aF(x)+ddx{bG(x)+cH(x)}by LIN with α=a,f(x)=F(x),β=1,and g(x)=bG(x)+cH(x)=aF(x)+bG(x)+cH(x)by LIN with α=b,f(x)=G(x),β=c,and g(x)=H(x)
This gives us linearity for three terms, namely (just replacing upper case names by lower case names)
ddx{af(x)+bg(x)+ch(x)}=af(x)+bg(x)+ch(x)
Just by repeating the above argument many times, we may generalise to linearity for n terms, for any natural number n:
ddx{a1f1(x)+a2f2(x)++anfn(x)}=a1f1(x)+a2f2(x)++anfn(x)
Similarly, while the product rule is stated for the product of two functions, it is not difficult to extend it to the product of three or more functions as the following example shows.

Example 2.6.6. Extending the product rule to more than two factors.

Once again, we’ll start by generalising the product rule to three factors.
ddx{F(x)G(x)H(x)}=F(x)G(x)H(x)+F(x)ddx{G(x)H(x)}by PR withf(x)=F(x) and g(x)=G(x)H(x)=F(x)G(x)H(x)+F(x){G(x)H(x)+G(x)H(x)}by PR withf(x)=G(x) and g(x)=H(x)
This gives us a product rule for three factors, namely (just replacing upper case names by lower case names)
ddx{f(x)g(x)h(x)}=f(x)g(x)h(x)+f(x)g(x)h(x)+f(x)g(x)h(x)
Observe that when we differentiate a product of three factors, the answer is a sum of three terms and in each term the derivative acts on exactly one of the original factors. Just by repeating the above argument many times, we may generalise the product rule to give the derivative of a product of n factors, for any natural number n:
ddx{f1(x)f2(x)fn(x)}=f1(x)f2(x)fn(x)+f1(x)f2(x)fn(x)+f1(x)f2(x)fn(x)
We can also write the above as
ddx{f1(x)f2(x)fn(x)}=[f1(x)f1(x)+f2(x)f2(x)++fn(x)fn(x)]f1(x)f2(x)fn(x)
When we differentiate a product of n factors, the answer is a sum of n terms and in each term the derivative acts on exactly one of the original factors. In the first term, the derivative acts on the first of the original factors. In the second term, the derivative acts on the second of the original factors. And so on.
If we make f1(x)=f2(x)==fn(x)=f(x) then each of the n terms on the right hand side of the above equation is the product of f(x) and exactly n1 f(x)’s, and so is exactly f(x)n1f(x). So we get the following useful result
ddxf(x)n=nf(x)n1f(x).
This last result is quite useful, so let us write it as a lemma for future reference.
This immediately gives us another useful result.

Example 2.6.8. Derivative of xn.

We can now compute the derivative of xn for any natural number n. Start with Lemma 2.6.7 and substitute f(x)=x and f(x)=1:
ddxxn=nxn11=nxn1
Again — this is a result we will come back to quite a few times in the future, so we should make sure we can refer to it easily. However, at present this statement only holds when n is a positive integer. With a little more work we can extend this to compute xq where q is any positive rational number and then any rational number at all (positive or negative). So let us hold off for a little longer. Instead we can make it a lemma, since it will be an ingredient in quite a few of the examples following below and in constructing the final corollary.
Back to more examples.

Example 2.6.10. Derivative of a polynomial.

ddx{2x3+4x5}=2ddx{x3}+4ddx{x5}by LIN with α=2,f(x)=x3,β=4 and g(x)=x5=2{3x2}+4{5x4}

by Lemma 2.6.9, once with n=3, and once with n=5

=6x2+20x4

Example 2.6.11. Derivative of product of polynomials.

In this example we’ll compute ddx{(3x+9)(x2+4x3)} in two different ways. For the first, we’ll start with the product rule.
ddx{(3x+9)(x2+4x3)}={ddx(3x+9)} (x2+4x3)+(3x+9) ddx{x2+4x3}={3×1+9×0} (x2+4x3)+(3x+9) {2x+4(3x2)}=3(x2+4x3)+(3x+9) (2x+12x2)=3x2+12x3+(6x2+18x+36x3+108x2)=18x+117x2+48x3
For the second, we expand the product first and then differentiate.
ddx{(3x+9)(x2+4x3)}=ddx{9x2+39x3+12x4}=9(2x)+39(3x2)+12(4x3)=18x+117x2+48x3

Example 2.6.12. Derivative of a rational function.

ddx{4x37x4x2+1}=(12x27)(4x2+1)(4x37x)(8x)(4x2+1)2by QR with f(x)=4x37x,f(x)=12x27,and g(x)=4x2+1,g(x)=8x=(48x416x27)(32x456x2)(4x2+1)2=16x4+40x27(4x2+1)2

Example 2.6.13. Derivative of a cube-root.

In this example, we’ll use a little trickery to find the derivative of x3. The trickery consists of observing that, by the definition of the cube root,
x=(x3)3.
Since both sides of the expression are the same, they must have the same derivatives:
ddx{x}=ddx(x3)3.
We already know by Theorem 2.2.4 that
ddx{x}=1
and that, by Lemma 2.6.7 with n=3 and f(x)=x3,
ddx(x3)3=3 (x3)2ddx{x3}=3x2/3ddx{x3}.
Since we know that ddx{x}=ddx(x3)3, we must have
1=3x2/3ddx{x3}

which we can rearrange to give the result we need

ddx{x3}=13x2/3

Example 2.6.14. Derivative of a positive rational power of x.

In this example, we’ll use the same trickery as in Example 2.6.13 to find the derivative xp/q for any two natural numbers p and q. By definition of the qth root,
xp=(xp/q)q.
That is, xp and (xp/q)q are the same function, and so have the same derivative. So we differentiate both of them. We already know that, by Lemma 2.6.9 with n=p,
ddx{xp}=pxp1
and that, by Lemma 2.6.7 with n=q and f(x)=xp/q,
ddx{(xp/q)q}=q (xp/q)q1 ddx{xp/q}
Remember that (xa)b=x(ab). Now these two derivatives must be the same. So
pxp1=qx(pqp)/qddx{xp/q}

and, rearranging things,

ddx{xp/q}=pqxp1(pqp)/q=pqx(pqqpq+p)/q=pqxpq1
So finally
ddx{xp/q}=pqxpq1
Notice that this has the same form as Lemma 2.6.9, above, except with n=pq allowed to be any positive rational number, not just a positive integer.

Example 2.6.15. Derivative of xm.

In this example we’ll use the quotient rule to find the derivative of xm, for any natural number m.
By the special case of the quotient rule (Corollary 2.4.6) with g(x)=xm and g(x)=mxm1
ddx{xm}=ddx{1xm}=mxm1(xm)2=mxm1
Again, notice that this has the same form as Lemma 2.6.9, above, except with n=m being a negative integer.

Example 2.6.16. Derivative of a negative rational power of x.

In this example we’ll use the quotient rule to find the derivative of xp/q, for any pair of natural numbers p and q. By the special case the quotient rule (Corollary 2.4.6) with g(x)=xpq and g(x)=pqxpq1,
ddx{xpq}=ddx{1xpq}=pqxpq1(xpq)2=pq xpq1
Note that we have found, in Examples 2.2.2, 2.6.14 and 2.6.16, the derivative of xa for any rational number a, whether 0, positive, negative, integer or fractional. In all cases, the answer is
We shall show, in Example 2.10.5, that the formula ddxxa=axa1 in fact applies for all real numbers a, not just rational numbers.
Back in Example 2.2.9 we computed the derivative of x from the definition of the derivative. The above corollary (correctly) gives
ddxx12=12x12
but with far less work.
Here’s an (optional) messy example.

Example 2.6.18. Optional messy example.

Find the derivative of
f(x)=(x1)(2x)(1x2)x(3+2x)
  • As we seen before, the best strategy for dealing with nasty expressions is to break them up into easy pieces. We can think of f(x) as the five–fold product
    f(x)=f1(x)f2(x)f3(x)1f4(x)1f5(x)
    with
    f1(x)=x1f2(x)=2xf3(x)=1x2f4(x)=xf5(x)=3+2x
  • By now, the derivatives of the fj’s should be easy to find:
    f1(x)=12xf2(x)=1f3(x)=2xf4(x)=12xf5(x)=2
  • Now, to get the derivative f(x) we use the n–fold product rule which was developed in Example 2.6.6, together with the special case of the quotient rule (Corollary 2.4.6).
    f(x)=f1f2f31f41f5+f1f2f31f41f5+f1f2f31f41f5f1f2f3f4f421f5f1f2f31f4f5f52=[f1f1+f2f2+f3f3f4f4f5f5]f1f2f31f41f5=[12x(x1)12x2x1x212x23+2x](x1)(2x)(1x2)x(3+2x)
    The trick that we used in going from the first line to the second line, namely multiplying term number j by fj(x)fj(x) is often useful in simplifying the derivative of a product of many factors
     1 
    Also take a look at “logarithmic differentiation” in Section 2.10.
    .

Exercises 2.6.2 Exercises

Exercises — Stage 1 .

1.
Spot and correct the error(s) in the following calculation.
f(x)=2xx+1f(x)=2(x+1)+2x(x+1)2=2(x+1)(x+1)2=2x+1
2.
True or false: ddx{2x}=x2x1.

Exercises — Stage 2 .

3.
Differentiate f(x)=23x6+5x4+12x2+9 and factor the result.
4.
Differentiate s(t)=3t4+5t31t.
5.
Differentiate x(y)=(2y+1y)y3.
6.
Differentiate T(x)=x+1x2+3.
7. (✳).
Compute the derivative of (7x+2x2+3).
8.
What is f(0), when f(x)=(3x3+4x2+x+1)(2x+5)?
9.
Differentiate f(x)=3x3+1x2+5x.
10. (✳).
Compute the derivative of (3x2+52x)
11. (✳).
Compute the derivative of (2x23x2+5).
12. (✳).
Compute the derivative of (2x3+1x+2).
13. (✳).
For what values of x does the derivative of x1x2 exist? Explain your answer.
14.
Differentiate f(x)=(3x5+15x3+8)(3x2+8x5).
15.
Differentiate f(x)=(x2+5x+1)(x+x3)x.
16.
Find all x-values where f(x)=115x5+x453x3 has a horizontal tangent line.

Exercises — Stage 3 .

17. (✳).
Find an equation of a line that is tangent to both of the curves y=x2 and y=x22x+2 (at different points).
18.
[1998H] Find all lines that are tangent to both of the curves y=x2 and y=x2+2x5. Illustrate your answer with a sketch.
19. (✳).
Evaluate limx2(x201522015x2).