Chapter 3 – Derivatives of radical expressions =

Chapter 3 – Derivatives of radical expressions
π‘Ž
𝑏
π‘₯ βˆ’π‘ =
π‘₯𝑏 = π‘₯π‘Ž
1
π‘₯𝑐
1.) Write as a fractional exponent:
a)
π‘₯3
b)
4
π‘₯3
3
𝑐) π‘₯ 5
d)
7
π‘₯
2.) Write as a fractional exponent and simplify the term so that it is not a fraction:
π‘Ž)
1
π‘₯7
3
b)
1+π‘₯ 4
𝑐)
1
d)
5π‘₯ 3 π‘₯ 2 βˆ’1
6π‘₯
4
( 3π‘₯ 2 )
π‘₯ 2 +π‘₯βˆ’7
3.) Write as a radical:
7
a) π‘₯ 2
2
b) π‘₯ 5
3
5
c) π‘₯ βˆ’ 4
d) π‘₯ βˆ’ 3
4.) Write fractional exponents as a radical and make all integer exponents positive:
1
a) 5π‘₯ βˆ’3 (π‘₯ 2 + π‘₯ βˆ’ 3)2
1
b) 3π‘₯ 2 (π‘₯ 4 βˆ’ 1)βˆ’2
c) 2π‘₯
βˆ’2
1
(1 + π‘₯ 4 )βˆ’2 6βˆ’1 ( π‘₯ )
5.) Find the derivative of each of the following functions:
5
2
a)
f (x) = π‘₯ 3 βˆ’ π‘₯ 3
b)
f (t) =
c)
f (h) = 2 β„Ž + 3β„Ž
d)
f (c) = c 𝑐 +
e)
f (x) =
f)
f (b) =
g)
f (x) =
5
𝑑 + 4 𝑑5
5
𝑐2
π‘₯ 2 +4π‘₯+3
π‘₯
𝑏+𝑏
𝑏2
π‘₯ +
1
3
π‘₯
2
Answer key
π‘Ž
Chapter 3 – Derivatives of radical expressions
𝑏
π‘₯ βˆ’π‘ =
π‘₯𝑏 = π‘₯π‘Ž
1
π‘₯𝑐
1.) Write as a fractional exponent:
a)
π‘₯3
4
b)
3
3
π‘₯3
𝑐) π‘₯ 5
5
3
π‘Ž) π‘₯ 2
d)
π‘₯
1
c) π‘₯ 3
b) π‘₯ 4
7
d) π‘₯ 7
2.) Write as a fractional exponent and simplify the term so that it is not a fraction:
1
π‘Ž)
π‘₯7
7
π‘Ž) π‘₯ βˆ’ 2
3
b)
𝑐)
1+π‘₯ 4
b) 3 1 + π‘₯ 4
1
2
βˆ’
1
d)
5π‘₯ 3 π‘₯ 2 βˆ’1
c ) 5βˆ’1 π‘₯ βˆ’3 π‘₯ 2 βˆ’ 1
βˆ’
1
2
6π‘₯
4
( 3π‘₯ 2 )
π‘₯ 2 +π‘₯βˆ’7
d ) 2π‘₯ βˆ’1 π‘₯ 2 + π‘₯ βˆ’ 7
3.) Write as a radical:
7
a) π‘₯ 2
a)
2
π‘₯7
2
3
5
b)
5
c) π‘₯ βˆ’ 4
b) π‘₯ 5
c)
π‘₯2
d) π‘₯ βˆ’ 3
1
4
d)
π‘₯3
1
3
π‘₯5
4.) Write fractional exponents as a radical and make all integer exponents positive:
1
a) 5π‘₯ βˆ’3 (π‘₯ 2 + π‘₯ βˆ’ 3)2
a)
5 π‘₯ 2 +π‘₯ βˆ’ 3
π‘₯3
1
b) 3π‘₯ 2 (π‘₯ 4 βˆ’ 1)βˆ’2
b)
3π‘₯ 2
π‘₯4 βˆ’ 1
c) 2π‘₯
c)
βˆ’2
1
(1 + π‘₯ 4 )βˆ’2 6βˆ’1 ( π‘₯ )
1
24π‘₯ 1+ π‘₯ 4
βˆ’
1
4
Answer key
5.) Find the derivative of each of the following functions:
5
a)
3
5βˆ™ π‘₯ 2
fβ€˜(x)=
b)
f (t) =
5
fβ€˜(t)=
c)
2
f (x) = π‘₯ 3 βˆ’ π‘₯ 3
3
1
5
5βˆ™
𝑑4
+ 10𝑑 𝑑
f (h) = 2 β„Ž + 3β„Ž
3
f (x) =
f (b) =
f (x) =
2
+
2
5
5βˆ™ 𝑐 3
π‘₯
3 π‘₯
2
2
+
π‘₯
3
βˆ’
2π‘₯ π‘₯
𝑏+𝑏
𝑏2
fβ€˜(b)=
g)
𝑐2
π‘₯ 2 +4π‘₯+3
fβ€˜(x)=
f)
3 𝑐
2 β„Ž
5
f (c) = c 𝑐 +
fβ€˜(c)=
e)
3βˆ™ 3 π‘₯
𝑑 + 4 𝑑5
fβ€˜(h)= 2+
d)
2
βˆ’
βˆ’3
2𝑏2 βˆ™
π‘₯ +
fβ€˜ (x ) =1+
𝑏
1
3
1
6
βˆ’
3 π‘₯5
1
𝑏2
2
π‘₯
βˆ’
2
3π‘₯ βˆ™
3
π‘₯2