Find the Derivative by Definition [ edit ] Find the derivative of the following functions using the limit definition of the derivative.
1.
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
2
x
{\displaystyle 2x}
2.
f
(
x
)
=
2
x
+
2
{\displaystyle f(x)=2x+2}
2
{\displaystyle 2}
3.
f
(
x
)
=
x
2
2
{\displaystyle f(x)={\frac {x^{2}}{2}}}
x
{\displaystyle x}
4.
f
(
x
)
=
2
x
2
+
4
x
+
4
{\displaystyle f(x)=2x^{2}+4x+4}
4
x
+
4
{\displaystyle 4x+4}
5.
f
(
x
)
=
x
+
2
{\displaystyle f(x)={\sqrt {x+2}}}
1
2
x
+
2
{\displaystyle {\frac {1}{2{\sqrt {x+2}}}}}
6.
f
(
x
)
=
1
x
{\displaystyle f(x)={\frac {1}{x}}}
−
1
x
2
{\displaystyle -{\frac {1}{x^{2}}}}
7.
f
(
x
)
=
3
x
+
1
{\displaystyle f(x)={\frac {3}{x+1}}}
−
3
(
x
+
1
)
2
{\displaystyle {\frac {-3}{(x+1)^{2}}}}
8.
f
(
x
)
=
1
x
+
1
{\displaystyle f(x)={\frac {1}{\sqrt {x+1}}}}
−
1
2
(
x
+
1
)
3
/
2
{\displaystyle {\frac {-1}{2(x+1)^{3/2}}}}
9.
f
(
x
)
=
x
x
+
2
{\displaystyle f(x)={\frac {x}{x+2}}}
2
(
x
+
2
)
2
{\displaystyle {\frac {2}{(x+2)^{2}}}}
Solutions
Prove the Constant Rule [ edit ]
10. Use the definition of the derivative to prove that for any fixed real number
c
{\displaystyle c}
,
d
d
x
[
c
⋅
f
(
x
)
]
=
c
⋅
d
d
x
[
f
(
x
)
]
{\displaystyle {\frac {d}{dx}}[c\cdot f(x)]=c\cdot {\frac {d}{dx}}[f(x)]}
d
d
x
[
c
⋅
f
(
x
)
]
=
lim
Δ
x
→
0
c
⋅
f
(
x
+
Δ
x
)
−
c
⋅
f
(
x
)
Δ
x
=
c
lim
Δ
x
→
0
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
=
c
d
d
x
[
f
(
x
)
]
{\displaystyle {\begin{aligned}{\frac {d}{dx}}[c\cdot f(x)]&=\lim _{\Delta x\to 0}{\frac {c\cdot f(x+\Delta x)-c\cdot f(x)}{\Delta x}}\\&=c\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\\&=c{\frac {d}{dx}}[f(x)]\end{aligned}}}
Solutions
Find the Derivative by Rules [ edit ] Find the derivative of the following functions:
Power Rule [ edit ]
11.
f
(
x
)
=
2
x
2
+
4
{\displaystyle f(x)=2x^{2}+4}
f
′
(
x
)
=
4
x
{\displaystyle f'(x)=4x}
12.
f
(
x
)
=
3
x
3
{\displaystyle f(x)=3{\sqrt[{3}]{x}}\,}
f
′
(
x
)
=
1
x
2
3
{\displaystyle f'(x)={\frac {1}{\sqrt[{3}]{x^{2}}}}}
13.
f
(
x
)
=
2
x
5
+
8
x
2
+
x
−
78
{\displaystyle f(x)=2x^{5}+8x^{2}+x-78}
f
′
(
x
)
=
10
x
4
+
16
x
+
1
{\displaystyle f'(x)=10x^{4}+16x+1}
14.
f
(
x
)
=
7
x
7
+
8
x
5
+
x
3
+
x
2
−
x
{\displaystyle f(x)=7x^{7}+8x^{5}+x^{3}+x^{2}-x}
f
′
(
x
)
=
49
x
6
+
40
x
4
+
3
x
2
+
2
x
−
1
{\displaystyle f'(x)=49x^{6}+40x^{4}+3x^{2}+2x-1}
15.
f
(
x
)
=
1
x
2
+
3
x
1
3
{\displaystyle f(x)={\frac {1}{x^{2}}}+3x^{\frac {1}{3}}}
f
′
(
x
)
=
−
2
x
3
+
1
x
2
3
{\displaystyle f'(x)={\frac {-2}{x^{3}}}+{\frac {1}{\sqrt[{3}]{x^{2}}}}}
16.
f
(
x
)
=
3
x
15
+
x
2
17
+
2
x
{\displaystyle f(x)=3x^{15}+{\frac {x^{2}}{17}}+{\frac {2}{\sqrt {x}}}}
f
′
(
x
)
=
45
x
14
+
2
x
17
−
1
x
x
{\displaystyle f'(x)=45x^{14}+{\frac {2x}{17}}-{\frac {1}{x{\sqrt {x}}}}}
17.
f
(
x
)
=
3
x
4
−
x
4
+
x
{\displaystyle f(x)={\frac {3}{x^{4}}}-{\sqrt[{4}]{x}}+x}
f
′
(
x
)
=
−
12
x
5
−
1
4
x
3
4
+
1
{\displaystyle f'(x)={\frac {-12}{x^{5}}}-{\frac {1}{4{\sqrt[{4}]{x^{3}}}}}+1}
18.
f
(
x
)
=
6
x
1
/
3
−
x
0.4
+
9
x
2
{\displaystyle f(x)=6x^{1/3}-x^{0.4}+{\frac {9}{x^{2}}}}
f
′
(
x
)
=
2
x
2
3
−
0.4
x
0.6
−
18
x
3
{\displaystyle f'(x)={\frac {2}{\sqrt[{3}]{x^{2}}}}-{\frac {0.4}{x^{0.6}}}-{\frac {18}{x^{3}}}}
19.
f
(
x
)
=
1
x
3
+
x
{\displaystyle f(x)={\frac {1}{\sqrt[{3}]{x}}}+{\sqrt {x}}}
f
′
(
x
)
=
−
1
3
x
4
3
+
1
2
x
{\displaystyle f'(x)={\frac {-1}{3{\sqrt[{3}]{x^{4}}}}}+{\frac {1}{2{\sqrt {x}}}}}
Solutions
Product Rule [ edit ]
20.
f
(
x
)
=
(
x
4
+
4
x
+
2
)
(
2
x
+
3
)
{\displaystyle f(x)=(x^{4}+4x+2)(2x+3)}
10
x
4
+
12
x
3
+
16
x
+
16
{\displaystyle 10x^{4}+12x^{3}+16x+16}
21.
f
(
x
)
=
(
2
x
−
1
)
(
3
x
2
+
2
)
{\displaystyle f(x)=(2x-1)(3x^{2}+2)}
18
x
2
−
6
x
+
4
{\displaystyle 18x^{2}-6x+4}
22.
f
(
x
)
=
(
x
3
−
12
x
)
(
3
x
2
+
2
x
)
{\displaystyle f(x)=(x^{3}-12x)(3x^{2}+2x)}
15
x
4
+
8
x
3
−
108
x
2
−
48
x
{\displaystyle 15x^{4}+8x^{3}-108x^{2}-48x}
23.
f
(
x
)
=
(
2
x
5
−
x
)
(
3
x
+
1
)
{\displaystyle f(x)=(2x^{5}-x)(3x+1)}
36
x
5
+
10
x
4
−
6
x
−
1
{\displaystyle 36x^{5}+10x^{4}-6x-1}
24.
f
(
x
)
=
(
5
x
2
+
3
)
(
2
x
+
7
)
{\displaystyle f(x)=(5x^{2}+3)(2x+7)}
30
x
2
+
70
x
+
6
{\displaystyle 30x^{2}+70x+6}
25.
f
(
x
)
=
3
x
2
(
5
x
2
+
1
)
4
{\displaystyle f(x)=3x^{2}(5x^{2}+1)^{4}}
6
x
(
25
x
2
+
1
)
(
5
x
2
+
1
)
3
{\displaystyle 6x(25x^{2}+1)(5x^{2}+1)^{3}}
26.
f
(
x
)
=
x
3
(
2
x
2
−
x
+
4
)
4
{\displaystyle f(x)=x^{3}(2x^{2}-x+4)^{4}}
x
2
(
2
x
2
−
x
+
4
)
3
(
22
x
2
−
7
x
+
12
)
{\displaystyle x^{2}(2x^{2}-x+4)^{3}(22x^{2}-7x+12)}
27.
f
(
x
)
=
5
x
2
(
x
3
−
x
+
1
)
3
{\displaystyle f(x)=5x^{2}(x^{3}-x+1)^{3}}
5
x
(
x
3
−
x
+
1
)
2
(
11
x
3
−
2
x
+
1
)
{\displaystyle 5x(x^{3}-x+1)^{2}(11x^{3}-2x+1)}
28.
f
(
x
)
=
(
2
−
x
)
6
(
5
+
2
x
)
4
{\displaystyle f(x)=(2-x)^{6}(5+2x)^{4}}
2
(
x
−
2
)
5
(
2
x
+
5
)
3
(
10
x
+
7
)
{\displaystyle 2(x-2)^{5}(2x+5)^{3}(10x+7)}
Solutions
Quotient Rule [ edit ]
24.
f
(
x
)
=
2
x
+
1
x
+
5
{\displaystyle f(x)={\frac {2x+1}{x+5}}}
f
′
(
x
)
=
9
(
x
+
5
)
2
{\displaystyle f'(x)={\frac {9}{(x+5)^{2}}}}
25.
f
(
x
)
=
3
x
4
+
2
x
+
2
3
x
2
+
1
{\displaystyle f(x)={\frac {3x^{4}+2x+2}{3x^{2}+1}}}
f
′
(
x
)
=
18
x
5
+
12
x
3
−
6
x
2
−
12
x
+
2
(
3
x
2
+
1
)
2
{\displaystyle f'(x)={\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}}}
26.
f
(
x
)
=
x
3
2
+
1
x
+
2
{\displaystyle f(x)={\frac {x^{\frac {3}{2}}+1}{x+2}}}
f
′
(
x
)
=
x
x
+
6
x
−
2
2
(
x
+
2
)
2
{\displaystyle f'(x)={\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}}}
27.
d
(
u
)
=
u
3
+
2
u
3
{\displaystyle d(u)={\frac {u^{3}+2}{u^{3}}}}
d
′
(
u
)
=
−
6
u
4
{\displaystyle d'(u)=-{\frac {6}{u^{4}}}}
28.
f
(
x
)
=
x
2
+
x
2
x
−
1
{\displaystyle f(x)={\frac {x^{2}+x}{2x-1}}}
f
′
(
x
)
=
2
x
2
−
2
x
−
1
(
2
x
−
1
)
2
{\displaystyle f'(x)={\frac {2x^{2}-2x-1}{(2x-1)^{2}}}}
29.
f
(
x
)
=
x
+
1
2
x
2
+
2
x
+
3
{\displaystyle f(x)={\frac {x+1}{2x^{2}+2x+3}}}
f
′
(
x
)
=
−
2
x
2
−
4
x
+
1
(
2
x
2
+
2
x
+
3
)
2
{\displaystyle f'(x)={\frac {-2x^{2}-4x+1}{(2x^{2}+2x+3)^{2}}}}
30.
f
(
x
)
=
16
x
4
+
2
x
2
x
{\displaystyle f(x)={\frac {16x^{4}+2x^{2}}{x}}}
f
′
(
x
)
=
48
x
2
+
2
{\displaystyle f'(x)=48x^{2}+2}
f
(
x
)
=
8
x
3
+
2
5
x
+
5
{\displaystyle f(x)={\frac {8x^{3}+2}{5x+5}}}
f
′
(
x
)
=
2
(
8
x
3
+
12
x
2
−
1
)
5
(
x
+
1
)
2
{\displaystyle f'(x)={\frac {2(8x^{3}+12x^{2}-1)}{5(x+1)^{2}}}}
f
(
x
)
=
(
3
x
−
2
)
2
x
{\displaystyle f(x)={\frac {(3x-2)^{2}}{\sqrt {x}}}}
f
′
(
x
)
=
(
3
x
−
2
)
(
9
x
+
2
)
2
x
x
{\displaystyle f'(x)={\frac {(3x-2)(9x+2)}{2x{\sqrt {x}}}}}
f
(
x
)
=
x
2
x
−
1
{\displaystyle f(x)={\frac {\sqrt {x}}{2x-1}}}
f
′
(
x
)
=
−
(
2
x
+
1
)
2
x
(
2
x
−
1
)
2
{\displaystyle f'(x)={\frac {-(2x+1)}{2{\sqrt {x}}(2x-1)^{2}}}}
f
(
x
)
=
4
x
−
3
x
+
2
{\displaystyle f(x)={\frac {4x-3}{x+2}}}
f
′
(
x
)
=
11
(
x
+
2
)
2
{\displaystyle f'(x)={\frac {11}{(x+2)^{2}}}}
f
(
x
)
=
4
x
+
3
2
x
−
1
{\displaystyle f(x)={\frac {4x+3}{2x-1}}}
f
′
(
x
)
=
−
10
(
2
x
−
1
)
2
{\displaystyle f'(x)={\frac {-10}{(2x-1)^{2}}}}
f
(
x
)
=
x
2
x
+
3
{\displaystyle f(x)={\frac {x^{2}}{x+3}}}
f
′
(
x
)
=
x
(
x
+
6
)
(
x
+
3
)
2
{\displaystyle f'(x)={\frac {x(x+6)}{(x+3)^{2}}}}
f
(
x
)
=
x
5
3
−
x
{\displaystyle f(x)={\frac {x^{5}}{3-x}}}
f
′
(
x
)
=
x
4
(
−
4
x
+
15
)
(
3
−
x
)
2
{\displaystyle f'(x)={\frac {x^{4}(-4x+15)}{(3-x)^{2}}}}
Solutions
Chain Rule [ edit ]
31.
f
(
x
)
=
(
x
+
5
)
2
{\displaystyle f(x)=(x+5)^{2}}
f
′
(
x
)
=
2
(
x
+
5
)
{\displaystyle f'(x)=2(x+5)}
32.
g
(
x
)
=
(
x
3
−
2
x
+
5
)
2
{\displaystyle g(x)=(x^{3}-2x+5)^{2}}
g
′
(
x
)
=
2
(
x
3
−
2
x
+
5
)
(
3
x
2
−
2
)
{\displaystyle g'(x)=2(x^{3}-2x+5)(3x^{2}-2)}
33.
f
(
x
)
=
1
−
x
2
{\displaystyle f(x)={\sqrt {1-x^{2}}}}
f
′
(
x
)
=
−
x
1
−
x
2
{\displaystyle f'(x)=-{\frac {x}{\sqrt {1-x^{2}}}}}
34.
f
(
x
)
=
(
2
x
+
4
)
3
4
x
3
+
1
{\displaystyle f(x)={\frac {(2x+4)^{3}}{4x^{3}+1}}}
f
′
(
x
)
=
6
(
4
x
3
+
1
)
(
2
x
+
4
)
2
−
(
2
x
+
4
)
3
(
12
x
2
)
(
4
x
3
+
1
)
2
{\displaystyle f'(x)={\frac {6(4x^{3}+1)(2x+4)^{2}-(2x+4)^{3}(12x^{2})}{(4x^{3}+1)^{2}}}}
35.
f
(
x
)
=
(
2
x
+
1
)
2
x
+
2
{\displaystyle f(x)=(2x+1){\sqrt {2x+2}}}
f
′
(
x
)
=
2
2
x
+
2
+
2
x
+
1
2
x
+
2
{\displaystyle f'(x)=2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}}
36.
f
(
x
)
=
2
x
+
1
2
x
+
2
{\displaystyle f(x)={\frac {2x+1}{\sqrt {2x+2}}}}
f
′
(
x
)
=
2
x
+
3
(
2
x
+
2
)
2
x
+
2
{\displaystyle f'(x)={\frac {2x+3}{(2x+2){\sqrt {2x+2}}}}}
37.
f
(
x
)
=
2
x
2
+
1
(
3
x
4
+
2
x
)
2
{\displaystyle f(x)={\sqrt {2x^{2}+1}}(3x^{4}+2x)^{2}}
f
′
(
x
)
=
2
x
(
3
x
4
+
2
x
)
2
2
x
2
+
1
+
2
x
2
+
1
(
2
)
(
3
x
4
+
2
x
)
(
12
x
3
+
2
)
{\displaystyle f'(x)={\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)}
38.
f
(
x
)
=
2
x
+
3
(
x
4
+
4
x
+
2
)
2
{\displaystyle f(x)={\frac {2x+3}{(x^{4}+4x+2)^{2}}}}
f
′
(
x
)
=
2
(
x
4
+
4
x
+
2
)
2
−
2
(
2
x
+
3
)
(
x
4
+
4
x
+
2
)
(
4
x
3
+
4
)
(
x
4
+
4
x
+
2
)
4
{\displaystyle f'(x)={\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}}
39.
f
(
x
)
=
x
3
+
1
(
x
2
−
1
)
{\displaystyle f(x)={\sqrt {x^{3}+1}}(x^{2}-1)}
f
′
(
x
)
=
3
x
(
x
2
−
1
)
2
x
3
+
1
+
2
x
x
3
+
1
{\displaystyle f'(x)={\frac {3x(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}}
40.
f
(
x
)
=
(
(
2
x
+
3
)
4
+
4
(
2
x
+
3
)
+
2
)
2
{\displaystyle f(x)=((2x+3)^{4}+4(2x+3)+2)^{2}}
f
′
(
x
)
=
2
(
(
2
x
+
3
)
4
+
4
(
2
x
+
3
)
+
2
)
(
8
(
2
x
+
3
)
3
+
8
)
{\displaystyle f'(x)=2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)}
41.
f
(
x
)
=
1
+
x
2
{\displaystyle f(x)={\sqrt {1+x^{2}}}}
f
′
(
x
)
=
x
1
+
x
2
{\displaystyle f'(x)={\frac {x}{\sqrt {1+x^{2}}}}}
Solutions
Exponentials [ edit ]
42.
f
(
x
)
=
(
3
x
2
+
e
)
e
2
x
{\displaystyle f(x)=(3x^{2}+e)e^{2x}}
f
′
(
x
)
=
6
x
e
2
x
+
2
e
2
x
(
3
x
2
+
e
)
{\displaystyle f'(x)=6xe^{2x}+2e^{2x}(3x^{2}+e)}
43.
f
(
x
)
=
e
2
x
2
+
3
x
{\displaystyle f(x)=e^{2x^{2}+3x}}
f
′
(
x
)
=
(
4
x
+
3
)
e
2
x
2
+
3
x
{\displaystyle f'(x)=(4x+3)e^{2x^{2}+3x}}
44.
f
(
x
)
=
e
e
2
x
2
+
1
{\displaystyle f(x)=e^{e^{2x^{2}+1}}}
f
′
(
x
)
=
4
x
e
2
x
2
+
1
+
e
2
x
2
+
1
{\displaystyle f'(x)=4xe^{2x^{2}+1+e^{2x^{2}+1}}}
45.
f
(
x
)
=
4
x
{\displaystyle f(x)=4^{x}}
f
′
(
x
)
=
ln
(
4
)
4
x
{\displaystyle f'(x)=\ln(4)4^{x}}
Solutions
Logarithms [ edit ]
46.
f
(
x
)
=
2
x
−
3
⋅
3
x
3
−
2
+
ln
(
x
)
{\displaystyle f(x)=2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+\ln(x)}
f
′
(
x
)
=
3
ln
(
2
)
2
x
−
3
x
3
−
2
+
9
x
2
2
x
−
4
x
3
−
2
+
1
x
{\displaystyle f'(x)=3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-4}}{\sqrt {x^{3}-2}}}+{\frac {1}{x}}}
47.
f
(
x
)
=
ln
(
x
)
−
2
e
x
+
x
{\displaystyle f(x)=\ln(x)-2e^{x}+{\sqrt {x}}}
f
′
(
x
)
=
1
x
−
2
e
x
+
1
2
x
{\displaystyle f'(x)={\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}}
48.
f
(
x
)
=
ln
(
ln
(
x
3
(
x
+
1
)
)
)
{\displaystyle f(x)=\ln(\ln(x^{3}(x+1)))}
f
′
(
x
)
=
4
x
3
+
3
x
2
x
3
(
x
+
1
)
ln
(
x
3
(
x
+
1
)
)
{\displaystyle f'(x)={\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}}}
49.
f
(
x
)
=
ln
(
2
x
2
+
3
x
)
{\displaystyle f(x)=\ln(2x^{2}+3x)}
f
′
(
x
)
=
4
x
+
3
2
x
2
+
3
x
{\displaystyle f'(x)={\frac {4x+3}{2x^{2}+3x}}}
50.
f
(
x
)
=
log
4
(
x
)
+
2
ln
(
x
)
{\displaystyle f(x)=\log _{4}(x)+2\ln(x)}
f
′
(
x
)
=
1
x
ln
(
4
)
+
2
x
{\displaystyle f'(x)={\frac {1}{x\ln(4)}}+{\frac {2}{x}}}
Solutions
Trigonometric functions [ edit ]
51.
f
(
x
)
=
3
e
x
−
4
cos
(
x
)
−
ln
(
x
)
4
{\displaystyle f(x)=3e^{x}-4\cos(x)-{\frac {\ln(x)}{4}}}
f
′
(
x
)
=
3
e
x
+
4
sin
(
x
)
−
1
4
x
{\displaystyle f'(x)=3e^{x}+4\sin(x)-{\frac {1}{4x}}}
52.
f
(
x
)
=
sin
(
x
)
+
cos
(
x
)
{\displaystyle f(x)=\sin(x)+\cos(x)}
f
′
(
x
)
=
cos
(
x
)
−
sin
(
x
)
{\displaystyle f'(x)=\cos(x)-\sin(x)}
Solutions
More Differentiation [ edit ]
53.
d
d
x
[
(
x
3
+
5
)
10
]
{\displaystyle {\frac {d}{dx}}[(x^{3}+5)^{10}]}
30
x
2
(
x
3
+
5
)
9
{\displaystyle 30x^{2}(x^{3}+5)^{9}}
54.
d
d
x
[
x
3
+
3
x
]
{\displaystyle {\frac {d}{dx}}[x^{3}+3x]}
3
x
2
+
3
{\displaystyle 3x^{2}+3}
55.
d
d
x
[
(
x
+
4
)
(
x
+
2
)
(
x
−
3
)
]
{\displaystyle {\frac {d}{dx}}[(x+4)(x+2)(x-3)]}
(
x
+
2
)
(
x
−
3
)
+
(
x
+
4
)
(
x
−
3
)
+
(
x
+
4
)
(
x
+
2
)
{\displaystyle (x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)}
56.
d
d
x
[
x
+
1
3
x
2
]
{\displaystyle {\frac {d}{dx}}[{\frac {x+1}{3x^{2}}}]}
−
x
+
2
3
x
3
{\displaystyle -{\frac {x+2}{3x^{3}}}}
57.
d
d
x
[
3
x
3
]
{\displaystyle {\frac {d}{dx}}[3x^{3}]}
9
x
2
{\displaystyle 9x^{2}}
58.
d
d
x
[
x
4
sin
(
x
)
]
{\displaystyle {\frac {d}{dx}}[x^{4}\sin(x)]}
4
x
3
sin
(
x
)
+
x
4
cos
(
x
)
{\displaystyle 4x^{3}\sin(x)+x^{4}\cos(x)}
59.
d
d
x
[
2
x
]
{\displaystyle {\frac {d}{dx}}[2^{x}]}
ln
(
2
)
2
x
{\displaystyle \ln(2)2^{x}}
60.
d
d
x
[
e
x
2
]
{\displaystyle {\frac {d}{dx}}[e^{x^{2}}]}
2
x
e
x
2
{\displaystyle 2xe^{x^{2}}}
61.
d
d
x
[
e
2
x
]
{\displaystyle {\frac {d}{dx}}[e^{2^{x}}]}
2
e
2
x
{\displaystyle 2e^{2^{x}}}
Solutions
Implicit Differentiation [ edit ] Use implicit differentiation to find y'
62.
x
3
+
y
3
=
x
y
{\displaystyle x^{3}+y^{3}=xy}
y
′
=
y
−
3
x
2
3
y
2
−
x
{\displaystyle y'={\frac {y-3x^{2}}{3y^{2}-x}}}
63.
(
2
x
+
y
)
4
+
3
x
2
+
3
y
2
=
x
y
+
1
{\displaystyle (2x+y)^{4}+3x^{2}+3y^{2}={\frac {x}{y}}+1}
y
′
=
y
−
8
y
2
(
2
x
+
y
)
3
−
6
x
y
2
4
y
2
(
2
x
+
y
)
3
+
6
y
3
+
x
{\displaystyle y'={\frac {y-8y^{2}(2x+y)^{3}-6xy^{2}}{4y^{2}(2x+y)^{3}+6y^{3}+x}}}
Solutions
Logarithmic Differentiation [ edit ] Use logarithmic differentiation to find
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
64.
y
=
x
(
1
−
x
3
4
)
{\displaystyle y=x({\sqrt[{4}]{1-x^{3}}})}
y
′
=
1
−
x
3
4
−
3
x
3
4
(
1
−
x
3
)
3
4
{\displaystyle y'={\sqrt[{4}]{1-x^{3}}}-{\frac {3x^{3}}{4(1-x^{3})^{\frac {3}{4}}}}}
65.
y
=
x
+
1
1
−
x
{\displaystyle y={\sqrt {x+1 \over 1-x}}\,}
y
′
=
1
2
x
+
1
1
−
x
(
1
x
+
1
+
1
1
−
x
)
{\displaystyle y'={\frac {1}{2}}{\sqrt {\frac {x+1}{1-x}}}\left({\frac {1}{x+1}}+{\frac {1}{1-x}}\right)}
66.
y
=
(
2
x
)
2
x
{\displaystyle y=(2x)^{2x}}
y
′
=
(
2
x
)
2
x
(
2
ln
(
2
x
)
+
2
)
{\displaystyle y'=(2x)^{2x}(2\ln(2x)+2)}
67.
y
=
(
x
3
+
4
x
)
3
x
+
1
{\displaystyle y=(x^{3}+4x)^{3x+1}}
y
′
=
(
x
3
+
4
x
)
3
x
+
1
(
3
ln
(
x
3
+
4
x
)
+
(
3
x
+
1
)
(
3
x
2
+
4
)
x
3
+
4
x
)
{\displaystyle y'=(x^{3}+4x)^{3x+1}(3\ln(x^{3}+4x)+{\frac {(3x+1)(3x^{2}+4)}{x^{3}+4x}})}
68.
y
=
(
6
x
)
cos
(
x
)
+
1
{\displaystyle y=(6x)^{\cos(x)+1}}
y
′
=
6
x
cos
(
x
)
+
1
(
−
sin
(
x
)
ln
(
6
x
)
+
cos
(
x
)
+
1
x
)
{\displaystyle y'=6x^{\cos(x)+1}(-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}})}
Solutions
Equation of Tangent Line [ edit ] For each function,
f
{\displaystyle f}
x
{\displaystyle x}
f
{\displaystyle f}
f
{\displaystyle f}
69.
f
(
x
)
=
x
3
3
+
x
2
+
5
,
(
3
,
23
)
{\displaystyle f(x)={\frac {x^{3}}{3}}+x^{2}+5,\qquad (3,23)}
a)
x
=
0
,
−
2
{\displaystyle x=0,-2}
y
=
15
x
−
22
{\displaystyle y=15x-22}
70.
f
(
x
)
=
x
3
−
3
x
+
1
,
(
1
,
−
1
)
{\displaystyle f(x)=x^{3}-3x+1,\qquad (1,-1)}
a)
x
=
±
1
{\displaystyle x=\pm 1}
y
=
−
1
{\displaystyle y=-1}
71.
f
(
x
)
=
2
x
3
3
+
x
2
−
12
x
+
6
,
(
0
,
6
)
{\displaystyle f(x)={\frac {2x^{3}}{3}}+x^{2}-12x+6,\qquad (0,6)}
a)
x
=
2
,
−
3
{\displaystyle x=2,-3}
y
=
−
12
x
+
6
{\displaystyle y=-12x+6}
72.
f
(
x
)
=
2
x
+
1
x
,
(
1
,
3
)
{\displaystyle f(x)=2x+{\frac {1}{\sqrt {x}}},\qquad (1,3)}
a)
x
=
2
−
4
3
{\displaystyle x=2^{-{\frac {4}{3}}}}
y
=
3
x
2
+
3
2
{\displaystyle y={\frac {3x}{2}}+{\frac {3}{2}}}
73.
f
(
x
)
=
(
x
2
+
1
)
(
2
−
x
)
,
(
2
,
0
)
{\displaystyle f(x)=(x^{2}+1)(2-x),\qquad (2,0)}
a)
x
=
1
,
1
3
{\displaystyle x=1,{\frac {1}{3}}}
y
=
−
5
x
+
10
{\displaystyle y=-5x+10}
74.
f
(
x
)
=
2
x
3
3
+
5
x
2
2
+
2
x
+
1
,
(
3
,
95
2
)
{\displaystyle f(x)={\frac {2x^{3}}{3}}+{\frac {5x^{2}}{2}}+2x+1,\qquad (3,{\frac {95}{2}})}
a)
x
=
−
1
2
,
−
2
{\displaystyle x=-{\frac {1}{2}},-2}
y
=
35
x
−
115
2
{\displaystyle y=35x-{\frac {115}{2}}}
75. Find an equation of the tangent line to the graph defined by
(
x
−
y
−
1
)
3
=
x
{\displaystyle (x-y-1)^{3}=x}
at the point (1,-1).
y
=
2
x
3
−
5
3
{\displaystyle y={\frac {2x}{3}}-{\frac {5}{3}}}
76. Find an equation of the tangent line to the graph defined by
e
x
y
+
x
2
=
y
2
{\displaystyle e^{xy}+x^{2}=y^{2}}
at the point (1,0).
y
=
−
2
x
+
2
{\displaystyle y=-2x+2}
Solutions
Higher Order Derivatives [ edit ]
77. What is the second derivative of
3
x
4
+
3
x
2
+
2
x
{\displaystyle 3x^{4}+3x^{2}+2x}
?
36
x
2
+
6
{\displaystyle 36x^{2}+6}
78. Use induction to prove that the (n+1)th derivative of a n-th order polynomial is 0.
Solutions
![{\displaystyle {\frac {d}{dx}}[c\cdot f(x)]=c\cdot {\frac {d}{dx}}[f(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e100ac3ba73695d9c36ad8ada81efba4c2a01129)
![{\displaystyle {\begin{aligned}{\frac {d}{dx}}[c\cdot f(x)]&=\lim _{\Delta x\to 0}{\frac {c\cdot f(x+\Delta x)-c\cdot f(x)}{\Delta x}}\\&=c\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\\&=c{\frac {d}{dx}}[f(x)]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c07443c9244e623f4190ad90f2f8dda344f84a57)
![{\displaystyle f(x)=3{\sqrt[{3}]{x}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7b861dfb68b951e23d614557d2cb071669b0851)
![f'(x)={\frac {1}{{\sqrt[ {3}]{x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94b10c8c5e4cbed34f88eb9719f071d312ee68a5)
![f'(x)={\frac {-2}{x^{3}}}+{\frac {1}{{\sqrt[ {3}]{x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f3c4021f5bbdf8fa04027a4fe4c907a8c10cc18)
![{\displaystyle f(x)={\frac {3}{x^{4}}}-{\sqrt[{4}]{x}}+x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff2d5e5287d1e1f9e9a3f66fe3d81364fbed337c)
![f'(x)={\frac {-12}{x^{5}}}-{\frac {1}{4{\sqrt[ {4}]{x^{3}}}}}+1](https://wikimedia.org/api/rest_v1/media/math/render/svg/05c20d3197b7b30c1a5a6492e0b36998f19c1dc0)
![f'(x)={\frac {2}{{\sqrt[ {3}]{x^{2}}}}}-{\frac {0.4}{x^{{0.6}}}}-{\frac {18}{x^{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/afda870d55f5f874bfde34936a0bafe945ccd502)
![{\displaystyle f(x)={\frac {1}{\sqrt[{3}]{x}}}+{\sqrt {x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa56b0f9f1e3f9dbd84b8e241f26eec58e0f966e)
![{\displaystyle f'(x)={\frac {-1}{3{\sqrt[{3}]{x^{4}}}}}+{\frac {1}{2{\sqrt {x}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/244d0f0859c7036984f53a7fd8ed474028a5afcc)
![{\displaystyle {\frac {d}{dx}}[(x^{3}+5)^{10}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/322d22bf641e5d6a35d63be10b39c7353404dfcd)
![{\displaystyle {\frac {d}{dx}}[x^{3}+3x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/052c718ffadbb9cb4954b7dd5791f055ff1ce77c)
![{\frac {d}{dx}}[(x+4)(x+2)(x-3)]](https://wikimedia.org/api/rest_v1/media/math/render/svg/160ae1c70a96f198eb4c1fd16d0ed2940ce4acfb)
![{\displaystyle {\frac {d}{dx}}[{\frac {x+1}{3x^{2}}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9320f75bc1f03266f545ee487bd178d212956699)
![{\displaystyle {\frac {d}{dx}}[3x^{3}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d330c0494fcf0513cd524464634f6360c2ed4c23)
![{\displaystyle {\frac {d}{dx}}[x^{4}\sin(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c54697e3e6e8b7be626f8ad40a67ef061ab14bde)
![{\displaystyle {\frac {d}{dx}}[2^{x}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70c4dd6b628068248913d496b20327629fc271d2)
![{\displaystyle {\frac {d}{dx}}[e^{x^{2}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70feadeaf0a4d91cf66a795d8d2f169e942ec523)
![{\displaystyle {\frac {d}{dx}}[e^{2^{x}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea1c7dffc918e0d7462c5f97f9a145db84a16597)
![{\displaystyle y=x({\sqrt[{4}]{1-x^{3}}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24d04c97ee04319439a96db44e9efa9e702123a0)
![{\displaystyle y'={\sqrt[{4}]{1-x^{3}}}-{\frac {3x^{3}}{4(1-x^{3})^{\frac {3}{4}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8930f128d8c9e1d48122565131e3b9f812a2cd73)
