# 1. Basic Results

Differentiation is a very powerful mathematical tool. This package reviews two rules which let us calculate the derivatives of products of functions and also of ratios of functions. The rules are given without any proof.

It is convenient to list here the derivatives of some simple functions:

y axn  sin (ax) cos (ax) eax  ln (ax)
dy/dx naxn − 1  a cos (ax) − a sin (ax) a eax  1/x

Also, recall the Sum Rule:

d/dx(u + v) = du/dx + dv/dx

This simply states that the deriviative of the sum of two (or more) functions is given by the sum of their derivatives.

It should also be recalled that derivatives commute with constants, i.e.

if  y = af(x)   then   dy/dx = adf/dx

- where a is any constant.

Exercise 1: Differentiate the following with respect to x using the rules described above.

(a)y = 4x2 + 3x − 5

Solution: To calculate the given function's derivative with respect to x, we need the sum rule and also the rule that:

d/dx(axn) = naxn−1

In the first term a = 4 and n = 2, in the second term a = 3 and n = 1 while the third term is a constant and has zero derivative. This yields:

 d/dx(4x2 + 3x − 5) = 2×4x2−1 + 1×31−1 + 0 = 8x1 + 3x0 = 8x + 3
(b)y = 4 sin (3x)

Solution: To differentiate the given function with respect to x we use the rule

d/dx(sin (ax)) = a cos (ax)

In this case, a = 3. We also take the derivative through the constant 4. This gives:

 dy/dx = d/dx(4 sin (3x)) = 4d/dx(sin (3x)) = 4 × 3 cos (3x)) = 12 cos (3x))
(c)y = e−2x

Solution: To differentiate the given function with respect to x we need the rule

d/dx(eax) = aeax

In this case, a = 2. This implies:

d/dx(e−2x) = −2 eax
(d)y = ln (x2)

Solution: To differentiate the given function, it is helpful to recall that log(AB) = log(A) − log(B) (see the module on Logarithms). So:

ln (x2) = ln (x) − ln (2)

The rule:

d/dxln (x) = 1x

- together with the sum rule gives:

 d/dx(ln (x) − ln (2)) = d/dx(ln (x)) − d/dx(ln (2)) = 1⁄x − 0 = 1⁄x

- since ln (2) is a constant and the derivative of a constant is 0.

Click on questions to reveal their solutions

Quiz 1: What is the derivative of y = (13) e3t − 3 cos (2t3) with respect to t?

(a)3 e3t − 2 cos (2t3)Incorrect - please try again!
(b)e3t + 2 sin (t)Incorrect - please try again!
(c)e3t + 2 sin (2t3)Correct - well done!
(d)e3t − 2 sin (2t3)Incorrect - please try again!

Explanation: To differentiatey = (13) e3t − 3 cos (2t3) with respect to t, we need the sum rule and the results:

d/dt(eat) = aeat, d/dt(cos (at)) = −a sin (at)

This gives:

 dy/dx = (1⁄3) × 3 e3t − 3 × (−2 ⁄ 3) sin (2t⁄3) = e3t + 2 sin (2t⁄3) = 1⁄x

since ln (2) is a constant and the derivative of a constant is 0.

# 2. The Product Rule

The product rule states that if u and v are both functions of x, and y is their product, then the derivative of y is given by:

 if   y = uv   then   dy/dx = udv/dx + vdu/dx

Here is a systematic procedure for applying the product rule:

1. Factorise y into y = uv.

2. Calculate the derivatives du/dx and dv/dx.

3. Insert these results into the product rule.

4. Finally, perform any possible simplifications.

Example 1: The product rule can be used to calculate the derivative of y = x2sin (x). First recognise that y may be written as y=uv, where u, v and their derivatives are given by:

 u = x2 v = sin (x) du/dx = 2x dv/dx = cos (x)

Inserting this into the product rule yields:

 dy/dx = udv/dx + vdu/dx = x2 × cos (x) + sin (x) × 2x = x2 cos (x) + 2x sin (x) = x(x cos (x) + 2 sin (x))

- where the common factor of x has been extracted.

Exercise 2: Use the product rule to differentiate the following products of functions with respect to x

(a)y = uv, if u=xm, and v=xn

Solution: The function y = xm × xn = xm + n (see the module on Powers). Thus the rule:

d/dx(axn) = naxn − 1

tells us that:

dy/dx = (m + n) axm + n − 1

This example allows us to practise the product rule. From y = xm × xn, the product rule yields:

 dy/dx = udv/dx + vdu/dx = xm × nxn − 1 + xn × mxm − 1 = nxm + n − 1 + mxm + n − 1 = (m+n) xm + n − 1

- which is indeed the expected result.

(b)y = uv, if u = 3x4, and v = e−2x

Solution: To differentiate y = 3x4e−2x with respect to x we may use the results:

d/dx(3x4) = 12x3 and d/dx (e−2x) = −2 e−2x

together with the product rule:

 dy/dx = udv/dx + vdu/dx = 3x4 × (−2e−2x) + e−2x × 12x3 = −6x4e−2x + 12x3 e − 2x = (−6x + 12) x3e − 2x = 6 (2 − x) x3 e−2x
(c)y = uv, if u = x3, and v = cos (x)

Solution: To differentiate y = x3cos (x) with respect to x we may use the results

d/dx(x3) = 3x2 and d/dx(cos (x)) = −sin (x)

together with the product rule:

 dy/dx = udv/dx + vdu/dx = x3 × (−sin (x)) + cos (x) × 3x2 = −x3 sin (x) + 3x2 cos (x) = x2[3 cos (x) − x sin (x)]
(d)y = uv, if u = ex, and v = ln (x)

Solution: To differentiate y = ex ln (x) with respect to x we may use the results:

d/dx(ex) = ex and d/dx(ln (x)) = 1x

and the product rule to obtain:

 dy/dx = udv/dx + vdu/dx = ex × 1⁄x + ln (x) × ex = [1⁄x + ln (x)] ex

Click on questions to reveal their solutions

Exercise 3: Use the product rule to differentiate the following with respect to x.

(a)y = x e2x

Solution: To differentiate the given function with respect to x we rewrite y as y = uv where:

 u = x and v = e2x du/dx = 1 and dv/dx = 2 e2x

Substituting this into the product rule yields:

 dy/dx = udv/dx + vdu/dx = x × 2 e2x + e2x × 1 = 2x e2x + e2x = (2x + 1) e2x
(b)y = sin (x) cos (2x)

Solution: To differentiate the given function with respect to x we rewrite y as y = uv where:

 u = sin (x) and v = cos (2x) du/dx = cos (x) and dv/dx = −2sin (x)

Substituting this into the product rule yields:

 dy/dx = udv/dx + vdu/dx = sin (x) × (−2sin (2x)) + cos (2x) × cos (x) = −2 sin (x) sin (2x) + cos (x) cos (2x)
(c)y = x ln (4x2)

Solution: To differentiate the given function with respect to x we rewrite y as y = uv where:

 u = x and v = ln (4 x2) du/dx = 1 and dv/dx = 2 ⁄ x

To obtain dvdx note that from the properties of logarithms ln (4x2) = ln (4) + 2ln (x), and recall that the derivative of ln (x) is 1x. Substituting this into the product rule yields:

 dy/dx = udv/dx + vdu/dx = x × 2 ⁄x + ln (4 x2) × 1 = 2 + ln (4 x2)
(d)y = x ln (x)

Solution: To differentiate the given function with respect to x we rewrite y as y = uv where:

 u = √x = x1 ⁄ 2 and v = ln (x) du/dx = 1 ⁄ 2 x−1 ⁄ 2 and dv/dx = 1⁄x

Substituting this into the product rule yields:

 dy/dx = udv/dx + vdu/dx = x1 ⁄ 2 × 1⁄x + ln (x) × 1 ⁄ 2 x−1 ⁄ 2 = x(1 ⁄ 2) − 1 + 1 ⁄ 2 x−1 ⁄ 2 ln (x) = x−1 ⁄ 2[1 + 1 ⁄ 2 ln (x)]

Click on questions to reveal their solutions

# 3. The Quotient Rule

The quotient rule states that if u and v are both functions of x and y, then:

 if  y=u/v  then  dy/dx = (v du/dx − u dv/dx )/v2

Example 2: Consider y = 1 ⁄ sin (x). The derivative may be found by writing y = uv, where:

 u = 1 and v = sin (x) du/dx = 0 and dv/dx = cos (x)

Inserting this into the quotient rule yields:

 dy/dx = sin (x) × 0 − 1 × cos (x)/sin2(x) = −cos (x)/sin2(x)

Example 3: Consider y = tan(x) = sin (x)/cos (x). The derivative may be found by writing y = uv, where:

 u = sin (x) and v = cos (x) du/dx = cos (x) and dv/dx = −sin (x)

Inserting this into the quotient rule yields:

 dy/dx = (v du/dx − u dv/dx )/v2 = cos (x) × cos (x) − sin (x) × (−sin (x))/cos2(x) = cos2 (x) + sin2 (x)/cos2 (x) = 1/cos2(x)  since  cos2 (x) + sin2 (x) = 1

Exercise 4: Use the quotient rule to differentiate the following functions with respect to x.

(a)y = uv, if u = eax and v = ebx

Solution: The function y = eax ⁄ ebx = e(ab) x (see the module on Powers). Hence its derivative with respective to x is:

dy/dx = (ab)e(ab)x

This example can also be used to practise the quotient rule.

 u = eax and v = ebx du/dx = aeax and dv/dx = bebx
 dy/dx = (v du/dx − u dv/dx )/v2 = ebx × aeax − eax × bebx/(ebx)2 = aeax + bx − beax + bx/e2bx = (a − b) e(a + b) x/e2bx = (a − b) e(a − b) x

- which is the expected result.

(b)y = uv, if u = x + 1 and v = x − 1

Solution: To differentiate this function y = uv, note that

 u = x+1 and v = x−1 du/dx = 1 and dv/dx = 1
 dy/dx = (v du/dx − u dv/dx )/v2 = (x − 1) × 1 − (x + 1) × 1/(x − 1)2 = x − 1 − x − 1/(x − 1)2 = −2/(x − 1)2

Click on questions to reveal their solutions

Exercise 5: Use the quotient rule to differentiate the following functions with respect to x.

(a)y = sin (x) ⁄ (x + 1)

Solution: To differentiate the given function, write y = uv where:

 u = sin (x) and v = x + 1 du/dx = cos (x) and dv/dx = 1

Thus, from the quotient rule:

 dy/dx = (v du/dx − u dv/dx )/v2 = (x + 1) × cos (x) − sin (x)×1/(x+1)2 = (x + 1) cos (x) − sin (x)/(x + 1)2
(b)y = sin (2x) ⁄ cos (2x)

Solution: To differentiate the given function, write y = uv where:

 u = sin (2x) and v = cos (2x) du/dx = 2 cos (x) and dv/dx = −2sin (x)

Thus, from the quotient rule:

 dy/dx = (v du/dx − u dv/dx )/v2 = cos (2x)×2 cos (2x) − sin (2x) × (−2 sin (2x))/cos2 (2x) = 2 cos2 (2x) + 2sin2 (2x)/cos2(2x) = 2 (cos2(2x) + sin2 (2x))/cos2 (2x) = 2/cos2(2x)
(c)y = (2x + 1) ⁄ (x − 2)

Solution: To differentiate the given function, write y = uv where:

 u = 2x + 1 and v = x − 2 du/dx = 2 and dv/dx = 1

Thus, from the quotient rule:

 dy/dx = (v du/dx − u dv/dx )/v2 = (x−2)×2 − (2x + 1)×1/(x − 2)2 = 2x − 4 − 2x − 1/(x−2)2 = −5/(x−2)2
(d)y = x3 / (3x + 2)

Solution: To differentiate the given function, write y = uv where:

 u = √x3 and v = 3x+2 du/dx = 3⁄2 x1⁄2 and dv/dx = 3

Thus, from the quotient rule:

 dy/dx = (v du/dx − u dv/dx )/v2 = (3x + 2)×(3 ⁄ 2)x1 ⁄ 2 − x3 ⁄ 2 × 3/(3x + 2)2 = (9 ⁄ 2)x3 ⁄ 2 + 3x1 ⁄ 2 − 3x3 ⁄ 2/(3x + 2)2 = (9 ⁄ 2 − 3)x3 ⁄ 2 + 3x1 ⁄ 2/(3x + 2)2 = (3 ⁄ 2)x3 ⁄ 2 + 3x1 ⁄ 2/(3x + 2)2 = 3x1 ⁄ 2(x+2)/2(3x+2)2

Click on questions to reveal their solutions

Quiz 2: Select the derivative of y = cot(t) with respect to t. (Hint: cot(t) = cos (t) ⁄ sin (t))

(a)−sin (t) ⁄ cos (t)Incorrect - please try again!
(b)−1 ⁄ sin2(t)Correct - well done!
(c)(cos2(t) − sin2(t)) ⁄ sin2(t)Incorrect - please try again!
(d)2 cos (t) sin (t) ⁄ sin2(t)Incorrect - please try again!

Explanation: The quotient rule may be used to differentiate y = cot(t) with respect to t. Writing y = uv, with u = cos (t) and v = sin (t), this gives:

 dy/dx = (v du/dx − u dv/dx )/v2 = sin (t) × (−sin (t)) − cos (t) × cos (t)/sin2(t) = −(cos2(t) + sin2(t))/sin2(t) = −1/sin2(t)

Exercise 6: Use the appropriate rule to differentiate the following functions with respect to the given variable.

(a)y = (z + 1) sin (3z) with respect to z

Solution: To differentiate the given function, we rewrite y = uv where:

 u = x+1 and v = sin (3z) du/dz = 1 and dv/dz = 3cos (3z)

Using the product rule:

 dy/dx = udv/dz + vdu/dz = (z + 1) × 3 cos (3z) + sin (3z) × 1 = 3 (z + 1) cos (3z) + sin (3z)
(b)y = 3 (w2 + 1) ⁄ (w + 1) with respect to w

Solution: To differentiate the given function, we rewrite y = uv where:

 u = 3 (w2 + 1) and v = w + 1 du/dw = 6w and dv/dw = 1

Using the quotient rule:

 dy/dx = (v du/dx − u dv/dx )/v2 = (w + 1) × 6w − 3(w2 + 1)×1/(w + 1)2 = 6w2 + 6w−3w2−3/(w+1)2 = 3w2 + 6w−3/(w + 1)2 = 3(w2 + 2w−1)/(w+1)2
(c)W = e2tln (3t) with respect to t

Solution: To differentiate the given function with respect to t, we rewrite W = uv where:

 u = e2t and v = ln (3t) = ln (3) + ln (t) du/dt = 2e2t and dv/dt = 1 ⁄ t

Using the product rule:

 dW/dt = udv/dt + vdu/dt = e2t × 1 ⁄ t + ln (3t) × 2 e2t = [1 ⁄ t + 2 ln (3t)] e2t

Click on questions to reveal their solutions

Quiz 3: The derivative, dydx, yields the rate of change of y with respect to x. Find the rate of change of y = x ⁄ (x + 1) with respect to x.

(a)− 1 ⁄ (x + 1)2Incorrect - please try again!
(b)0Incorrect - please try again!
(c)(2x + 1) ⁄ (x + 1)2Incorrect - please try again!
(d)1 ⁄ (x + 1)2Correct - well done!

Explanation: To differentiate y = x ⁄ (x + 1) with respect to x, we may use the quotient rule. For y = uv where:

 u = x and v = x + 1 du/dw = 1 and dv/dw = 1

Using the quotient rule:

 dy/dx = (v du/dx − u dv/dx )/v2 = (x + 1)×1 − x×1/(x+1)2 = 1/(x + 1)2

# 4. Quiz on the Product and Quotient Rule

1. What is the derivative with respect to x of y = x(ln (x) − 1)?
(a)ln (x) + 1x
(b)ln (x)
(c)1
(d)1x
2. Velocity is the derivative of position with respect to time. If the position, x, of a body is given by x = 3te2t in metres at time t in seconds, what is its velocity?
(a)(6t + 3) e2tm s−1
(b)(6t2 + 3) e2tm s−1
(c)3 + e2tm s−1
(d)(3t + 2) e2tm s−1
3. What is the rate of change of y = (x2 + 1) ⁄ (x2−1) with respect to x?
(a) 1
(b)x
(c)4x3 ⁄ (x2−1)
(d)−4x ⁄ (x2 − 1)2