Mixed Differentiation Problems 2 Answers 11 - 15
1 of 2
http://mathsfirst.massey.ac.nz/Calculus/MixDiff/ex8ans.html
Mixed Differentiation Problems 2 Answers 11 - 15
We assume you are familiar enough with the basics, Multiple Rule and Sum and Difference Rules to no
longer need to mention which is being used when several are used with simple expressions such as
polynomials.
11. y = 2xe5 − 2x
This problem is a product of a basic function and a composite function, so use the
Product Rule then the Chain Rule.
= 2x
e5 − 2x + e5 − 2x
= 2xe5 − 2x
2x
using the Product Rule
(5 − 2x) + e5 − 2x
= 2xe5 − 2x(−2) + e5 − 2x(2)
= e5 − 2x(2 − 4x)
2x
using the Chain Rule
using basic derivatives
simplifying
12. y =
This problem is best tackled by expressing the function with a negative exponent
before differentiating using the Chain Rule.
y=
= (cos 5x)−2
= −2(cos 5x)−3
(cos 5x)
= −2(cos 5x)−3(−sin 5x)
using the Chain Rule
(5x)
= −2(cos 5x)−3(−sin 5x)(5)
= 10(sin 5x)(cos 5x)−3
using the Chain Rule
using basic derivatives
simplifying
13. y = (4 − x)(3 + x2)(2x4 + 1)
This problem is a product of polynomials. One way to tackle this is to expand out the
brackets before differentiating. Alternatively use the Product Rule.
y = (4 − x)(3 + x2)(2x4 + 1) = (12 − 3x + 4x2 − x3)(2x4 + 1) = 24x4 − 6x5 + 8x6 − 2x7 +
12 − 3x + 4x2 − x3
y = 12 − 3x + 4x2 − x3 + 24x4 − 6x5 + 8x6 − 2x7
= −3 + 8x − 3x2 + 96x3 − 30x4 + 48x5 − 14x6 using basic derivatives
14. y = ln (4x + 3)x
28/5/2013 13:49
Mixed Differentiation Problems 2 Answers 11 - 15
2 of 2
http://mathsfirst.massey.ac.nz/Calculus/MixDiff/ex8ans.html
This problem is best tackled by using the properties of logs first to obtain a product of a
basic function and a composite function, then use the Product Rule and the Chain Rule.
y = ln (4x + 3)x = x ln (4x + 3)
=x
ln (4x + 3) + ln (4x + 3)
= x(4x + 3)−1
x
using the Product Rule
(4x + 3) + ln (4x + 3)
= x(4x + 3)−1(4) + ln (4x + 3)(1)
= 4x(4x + 3)−1 + ln (4x + 3)
using the Chain Rule
x
using basic derivatives
simplifying
15. y =
This problem is a quotient of a product of polynomials. One way is to expand the
product and express the whole as a composite function before differentiating using the
Chain Rule.
y=
=
= 3(10x − 3x2 − x3)−1
=
= −3(10x − 3x2 − x3)−2
(10x − 3x2 − x3)
= −3(10x − 3x2 − x3)−2(10 − 6x − 3x2)
=
by Chain Rule
by basic derivatives
simplifying
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28/5/2013 13:49