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 | ax^{n}_{ } | sin (ax) | cos (ax) | e^{ax}_{ } | ln (ax) |
---|---|---|---|---|---|
dydx | nax^{n − 1}_{ } | a cos (ax) | − a sin (ax) | a e^{ax}_{ } | 1x |
Also, recall the Sum Rule:
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.
- where a is any constant.
Exercise 1: Differentiate the following with respect to x using the rules described above.
Solution: To calculate the given function's derivative with respect to x, we need the sum rule and also the rule that:
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:
ddx(4x^{2} + 3x − 5) | = | 2×4x^{2−1} + 1×3^{1−1} + 0 |
= | 8x^{1} + 3x^{0} | |
= | 8x + 3 |
Solution: To differentiate the given function with respect to x we use the rule
In this case, a = 3. We also take the derivative through the constant 4. This gives:
dydx | = | ddx(4 sin (3x)) |
= | 4ddx(sin (3x)) | |
= | 4 × 3 cos (3x)) | |
= | 12 cos (3x)) |
Solution: To differentiate the given function with respect to x we need the rule
In this case, a = 2. This implies:
Solution: To differentiate the given function, it is helpful to recall that log(A ⁄ B) = log(A) − log(B) (see the module on Logarithms). So:
The rule:
- together with the sum rule gives:
ddx(ln (x) − ln (2)) | = | ddx(ln (x)) − ddx(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 = (^{1}⁄_{3}) e^{3t} − 3 cos (^{2t}⁄_{3}) with respect to t?
Explanation: To differentiatey = (^{1}⁄_{3}) e^{3t} − 3 cos (^{2t}⁄_{3}) with respect to t, we need the sum rule and the results:
This gives:
dydx | = | (^{1}⁄_{3}) × 3 e^{3t} − 3 × (−2 ⁄ 3) sin (^{2t}⁄_{3}) |
= | e^{3t} + 2 sin (^{2t}⁄_{3}) | |
= | ^{1}⁄_{x} |
since ln (2) is a constant and the derivative of a constant is 0.
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 dydx = udvdx + vdudx |
Here is a systematic procedure for applying the product rule:
1. Factorise y into y = uv.
2. Calculate the derivatives dudx and dvdx.
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 = x^{2}sin (x). First recognise that y may be written as y=uv, where u, v and their derivatives are given by:
u = x^{2} | v = sin (x) |
dudx = 2x | dvdx = cos (x) |
Inserting this into the product rule yields:
dydx | = | udvdx + vdudx |
= | x^{2} × cos (x) + sin (x) × 2x | |
= | x^{2} 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
Solution: The function y = x^{m} × x^{n} = x^{m + n} (see the module on Powers). Thus the rule:
tells us that:
This example allows us to practise the product rule. From y = x^{m} × x^{n}, the product rule yields:
dydx | = | udvdx + vdudx |
= | x^{m} × nx^{n − 1} + x^{n} × mx^{m − 1} | |
= | nx^{m + n − 1} + mx^{m + n − 1} | |
= | (m+n) x^{m + n − 1} |
- which is indeed the expected result.
Solution: To differentiate y = 3x^{4}e^{−2x} with respect to x we may use the results:
together with the product rule:
dydx | = | udvdx + vdudx |
= | 3x^{4} × (−2e^{−2x}) + e^{−2x} × 12x^{3} | |
= | −6x^{4}e^{−2x} + 12x^{3} e^{ − 2x} | |
= | (−6x + 12) x^{3}e^{ − 2x} | |
= | 6 (2 − x) x^{3} e^{−2x} |
Solution: To differentiate y = x^{3}cos (x) with respect to x we may use the results
together with the product rule:
dydx | = | udvdx + vdudx |
= | x^{3} × (−sin (x)) + cos (x) × 3x^{2} | |
= | −x^{3} sin (x) + 3x^{2} cos (x) | |
= | x^{2}[3 cos (x) − x sin (x)] |
Solution: To differentiate y = e^{x} ln (x) with respect to x we may use the results:
and the product rule to obtain:
dydx | = | udvdx + vdudx |
= | e^{x} × ^{1}⁄_{x} + ln (x) × e^{x} | |
= | [^{1}⁄_{x} + ln (x)] e^{x} |
Click on questions to reveal their solutions
Exercise 3: Use the product rule to differentiate the following with respect to x.
Solution: To differentiate the given function with respect to x we rewrite y as y = uv where:
u = x | and | v = e^{2x} |
dudx = 1 | and | dvdx = 2 e^{2x} |
Substituting this into the product rule yields:
dydx | = | udvdx + vdudx |
= | x × 2 e^{2x} + e^{2x} × 1 | |
= | 2x e^{2x} + e^{2x} | |
= | (2x + 1) e^{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) |
dudx = cos (x) | and | dvdx = −2sin (x) |
Substituting this into the product rule yields:
dydx | = | udvdx + vdudx |
= | sin (x) × (−2sin (2x)) + cos (2x) × cos (x) | |
= | −2 sin (x) sin (2x) + cos (x) cos (2x) |
Solution: To differentiate the given function with respect to x we rewrite y as y = uv where:
u = x | and | v = ln (4 x^{2}) |
dudx = 1 | and | dvdx = 2 ⁄ x |
To obtain dv⁄dx note that from the properties of logarithms ln (4x^{2}) = ln (4) + 2ln (x), and recall that the derivative of ln (x) is ^{1}⁄_{x}. Substituting this into the product rule yields:
dydx | = | udvdx + vdudx |
= | x × 2 ⁄x + ln (4 x^{2}) × 1 | |
= | 2 + ln (4 x^{2}) |
Solution: To differentiate the given function with respect to x we rewrite y as y = uv where:
u = √x = x^{1 ⁄ 2} | and | v = ln (x) |
dudx = 1 ⁄ 2 x^{−1 ⁄ 2} | and | dvdx = ^{1}⁄_{x} |
Substituting this into the product rule yields:
dydx | = | udvdx + vdudx |
= | x^{1 ⁄ 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
The quotient rule states that if u and v are both functions of x and y, then:
if y=uv then dydx = (v dudx − u dvdx )v^{2} |
Example 2: Consider y = 1 ⁄ sin (x). The derivative may be found by writing y = u ⁄ v, where:
u = 1 | and | v = sin (x) |
dudx = 0 | and | dvdx = cos (x) |
Inserting this into the quotient rule yields:
dydx | = | sin (x) × 0 − 1 × cos (x)sin^{2}(x) |
= | −cos (x)sin^{2}(x) |
Example 3: Consider y = tan(x) = sin (x)cos (x). The derivative may be found by writing y = u ⁄ v, where:
u = sin (x) | and | v = cos (x) |
dudx = cos (x) | and | dvdx = −sin (x) |
Inserting this into the quotient rule yields:
dydx | = | (v dudx − u dvdx )v^{2} |
= | cos (x) × cos (x) − sin (x) × (−sin (x))cos^{2}(x) | |
= | cos^{2} (x) + sin^{2} (x)cos^{2} (x) | |
= | 1cos^{2}(x) since cos^{2} (x) + sin^{2} (x) = 1 |
Exercise 4: Use the quotient rule to differentiate the following functions with respect to x.
Solution: The function y = e^{ax} ⁄ e^{bx} = e^{(a − b) x} (see the module on Powers). Hence its derivative with respective to x is:
This example can also be used to practise the quotient rule.
u = e^{ax} | and | v = e^{bx} |
dudx = ae^{ax} | and | dvdx = be^{bx} |
dydx | = | (v dudx − u dvdx )v^{2} |
= | e^{bx} × ae^{ax} − e^{ax} × be^{bx}(e^{bx})^{2} | |
= | ae^{ax + bx} − be^{ax + bx}e^{2bx} | |
= | (a − b) e^{(a + b) x}e^{2bx} | |
= | (a − b) e^{(a − b) x} |
- which is the expected result.
Solution: To differentiate this function y = u ⁄ v, note that
u = x+1 | and | v = x−1 |
dudx = 1 | and | dvdx = 1 |
dydx | = | (v dudx − u dvdx )v^{2} |
= | (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.
Solution: To differentiate the given function, write y = u ⁄ v where:
u = sin (x) | and | v = x + 1 |
dudx = cos (x) | and | dvdx = 1 |
Thus, from the quotient rule:
dydx | = | (v dudx − u dvdx )v^{2} |
= | (x + 1) × cos (x) − sin (x)×1(x+1)^{2} | |
= | (x + 1) cos (x) − sin (x)(x + 1)^{2} |
Solution: To differentiate the given function, write y = u ⁄ v where:
u = sin (2x) | and | v = cos (2x) |
dudx = 2 cos (x) | and | dvdx = −2sin (x) |
Thus, from the quotient rule:
dydx | = | (v dudx − u dvdx )v^{2} |
= | cos (2x)×2 cos (2x) − sin (2x) × (−2 sin (2x))cos^{2} (2x) | |
= | 2 cos^{2} (2x) + 2sin^{2} (2x)cos^{2}(2x) | |
= | 2 (cos^{2}(2x) + sin^{2} (2x))cos^{2} (2x) | |
= | 2cos^{2}(2x) |
Solution: To differentiate the given function, write y = u ⁄ v where:
u = 2x + 1 | and | v = x − 2 |
dudx = 2 | and | dvdx = 1 |
Thus, from the quotient rule:
dydx | = | (v dudx − u dvdx )v^{2} |
= | (x−2)×2 − (2x + 1)×1(x − 2)^{2} | |
= | 2x − 4 − 2x − 1(x−2)^{2} | |
= | −5(x−2)^{2} |
Solution: To differentiate the given function, write y = u ⁄ v where:
u = √x^{3} | and | v = 3x+2 |
dudx = 3⁄2 x^{1⁄2} | and | dvdx = 3 |
Thus, from the quotient rule:
dydx | = | (v dudx − u dvdx )v^{2} |
= | (3x + 2)×(3 ⁄ 2)x^{1 ⁄ 2} − x^{3 ⁄ 2} × 3(3x + 2)^{2} | |
= | (9 ⁄ 2)x^{3 ⁄ 2} + 3x^{1 ⁄ 2} − 3x3 ⁄ 2(3x + 2)^{2} | |
= | (9 ⁄ 2 − 3)x^{3 ⁄ 2} + 3x^{1 ⁄ 2}(3x + 2)^{2} | |
= | (3 ⁄ 2)x^{3 ⁄ 2} + 3x^{1 ⁄ 2}(3x + 2)^{2} | |
= | 3x^{1 ⁄ 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))
Explanation: The quotient rule may be used to differentiate y = cot(t) with respect to t. Writing y = u ⁄ v, with u = cos (t) and v = sin (t), this gives:
dydx | = | (v dudx − u dvdx )v^{2} |
= | sin (t) × (−sin (t)) − cos (t) × cos (t)sin^{2}(t) | |
= | −(cos^{2}(t) + sin^{2}(t))sin^{2}(t) | |
= | −1sin^{2}(t) |
Exercise 6: Use the appropriate rule to differentiate the following functions with respect to the given variable.
Solution: To differentiate the given function, we rewrite y = uv where:
u = x+1 | and | v = sin (3z) |
dudz = 1 | and | dvdz = 3cos (3z) |
Using the product rule:
dydx | = | udvdz + vdudz |
= | (z + 1) × 3 cos (3z) + sin (3z) × 1 | |
= | 3 (z + 1) cos (3z) + sin (3z) |
Solution: To differentiate the given function, we rewrite y = u ⁄ v where:
u = 3 (w^{2} + 1) | and | v = w + 1 |
dudw = 6w | and | dvdw = 1 |
Using the quotient rule:
dydx | = | (v dudx − u dvdx )v^{2} |
= | (w + 1) × 6w − 3(w^{2} + 1)×1(w + 1)^{2} | |
= | 6w^{2} + 6w−3w^{2}−3(w+1)^{2} | |
= | 3w^{2} + 6w−3(w + 1)^{2} | |
= | 3(w^{2} + 2w−1)(w+1)^{2} |
Solution: To differentiate the given function with respect to t, we rewrite W = uv where:
u = e^{2t} | and | v = ln (3t) = ln (3) + ln (t) |
dudt = 2e^{2t} | and | dvdt = 1 ⁄ t |
Using the product rule:
dWdt | = | udvdt + vdudt |
= | e^{2t} × 1 ⁄ t + ln (3t) × 2 e^{2t} | |
= | [1 ⁄ t + 2 ln (3t)] e^{2t} |
Click on questions to reveal their solutions
Quiz 3: The derivative, dy⁄dx, 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.
Explanation: To differentiate y = x ⁄ (x + 1) with respect to x, we may use the quotient rule. For y = u ⁄ v where:
u = x | and | v = x + 1 |
dudw = 1 | and | dvdw = 1 |
Using the quotient rule:
dydx | = | (v dudx − u dvdx )v^{2} |
= | (x + 1)×1 − x×1(x+1)^{2} | |
= | 1(x + 1)^{2} |