Find the Derivative by Definition
edit
Prove the Constant Rule
edit
Find the Derivative by Rules
edit
Power Rule
edit
Product Rule
edit
Quotient Rule
edit
Chain Rule
edit
43.
f
(
x
)
=
(
x
+
5
)
2
{\displaystyle f(x)=(x+5)^{2}\,}
Let
f
(
x
)
=
g
(
h
(
x
)
)
;
g
(
x
)
=
x
2
;
h
(
x
)
=
x
+
5
{\displaystyle f(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x+5}
. Then
f
′
(
x
)
=
d
g
d
h
d
h
d
x
=
2
(
x
+
5
)
(
1
)
=
2
(
x
+
5
)
{\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x+5)(1)=\mathbf {2(x+5)} }
Let
f
(
x
)
=
g
(
h
(
x
)
)
;
g
(
x
)
=
x
2
;
h
(
x
)
=
x
+
5
{\displaystyle f(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x+5}
. Then
f
′
(
x
)
=
d
g
d
h
d
h
d
x
=
2
(
x
+
5
)
(
1
)
=
2
(
x
+
5
)
{\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x+5)(1)=\mathbf {2(x+5)} }
44.
g
(
x
)
=
(
x
3
−
2
x
+
5
)
2
{\displaystyle g(x)=(x^{3}-2x+5)^{2}\,}
Let
g
(
x
)
=
r
(
s
(
x
)
)
;
r
(
x
)
=
x
2
;
s
(
x
)
=
x
3
−
2
x
+
5
{\displaystyle g(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=x^{3}-2x+5}
. Then
g
′
(
x
)
=
d
r
d
s
d
s
d
x
=
2
(
x
3
−
2
x
+
5
)
(
3
x
2
−
2
)
{\displaystyle g'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=\mathbf {2(x^{3}-2x+5)(3x^{2}-2)} }
Let
g
(
x
)
=
r
(
s
(
x
)
)
;
r
(
x
)
=
x
2
;
s
(
x
)
=
x
3
−
2
x
+
5
{\displaystyle g(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=x^{3}-2x+5}
. Then
g
′
(
x
)
=
d
r
d
s
d
s
d
x
=
2
(
x
3
−
2
x
+
5
)
(
3
x
2
−
2
)
{\displaystyle g'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=\mathbf {2(x^{3}-2x+5)(3x^{2}-2)} }
45.
f
(
x
)
=
1
−
x
2
{\displaystyle f(x)={\sqrt {1-x^{2}}}\,}
Let
f
(
x
)
=
g
(
h
(
x
)
)
;
g
(
x
)
=
x
;
h
(
x
)
=
1
−
x
2
{\displaystyle f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1-x^{2}}
. Then
f
′
(
x
)
=
d
g
d
h
d
h
d
x
=
1
2
1
−
x
2
(
−
2
x
)
=
−
x
1
−
x
2
{\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1-x^{2}}}}}(-2x)=\mathbf {-{\frac {x}{\sqrt {1-x^{2}}}}} }
Let
f
(
x
)
=
g
(
h
(
x
)
)
;
g
(
x
)
=
x
;
h
(
x
)
=
1
−
x
2
{\displaystyle f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1-x^{2}}
. Then
f
′
(
x
)
=
d
g
d
h
d
h
d
x
=
1
2
1
−
x
2
(
−
2
x
)
=
−
x
1
−
x
2
{\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1-x^{2}}}}}(-2x)=\mathbf {-{\frac {x}{\sqrt {1-x^{2}}}}} }
46.
f
(
x
)
=
(
2
x
+
4
)
3
4
x
3
+
1
{\displaystyle f(x)={\frac {(2x+4)^{3}}{4x^{3}+1}}\,}
47.
f
(
x
)
=
(
2
x
+
1
)
2
x
+
2
{\displaystyle f(x)=(2x+1){\sqrt {2x+2}}\,}
48.
f
(
x
)
=
2
x
+
1
2
x
+
2
{\displaystyle f(x)={\frac {2x+1}{\sqrt {2x+2}}}\,}
49.
f
(
x
)
=
2
x
2
+
1
(
3
x
4
+
2
x
)
2
{\displaystyle f(x)={\sqrt {2x^{2}+1}}(3x^{4}+2x)^{2}\,}
50.
f
(
x
)
=
2
x
+
3
(
x
4
+
4
x
+
2
)
2
{\displaystyle f(x)={\frac {2x+3}{(x^{4}+4x+2)^{2}}}\,}
51.
f
(
x
)
=
x
3
+
1
(
x
2
−
1
)
{\displaystyle f(x)={\sqrt {x^{3}+1}}(x^{2}-1)\,}
52.
f
(
x
)
=
(
(
2
x
+
3
)
4
+
4
(
2
x
+
3
)
+
2
)
2
{\displaystyle f(x)=((2x+3)^{4}+4(2x+3)+2)^{2}\,}
53.
f
(
x
)
=
1
+
x
2
{\displaystyle f(x)={\sqrt {1+x^{2}}}\,}
Let
f
(
x
)
=
g
(
h
(
x
)
)
;
g
(
x
)
=
x
;
h
(
x
)
=
1
+
x
2
{\displaystyle f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1+x^{2}}
. Then
f
′
(
x
)
=
d
g
d
h
d
h
d
x
=
1
2
1
+
x
2
(
2
x
)
=
x
1
+
x
2
{\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}(2x)=\mathbf {\frac {x}{\sqrt {1+x^{2}}}} }
Let
f
(
x
)
=
g
(
h
(
x
)
)
;
g
(
x
)
=
x
;
h
(
x
)
=
1
+
x
2
{\displaystyle f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1+x^{2}}
. Then
f
′
(
x
)
=
d
g
d
h
d
h
d
x
=
1
2
1
+
x
2
(
2
x
)
=
x
1
+
x
2
{\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}(2x)=\mathbf {\frac {x}{\sqrt {1+x^{2}}}} }
Exponentials
edit
55.
f
(
x
)
=
e
2
x
2
+
3
x
{\displaystyle f(x)=e^{2x^{2}+3x}}
Let
f
(
x
)
=
g
(
h
(
x
)
)
;
g
(
x
)
=
e
x
;
h
(
x
)
=
2
x
2
+
3
x
{\displaystyle f(x)=g(h(x));\quad g(x)=e^{x};\quad h(x)=2x^{2}+3x}
. Then
f
′
(
x
)
=
d
g
d
h
d
h
d
x
=
e
2
x
2
+
3
x
(
4
x
+
3
)
=
(
4
x
+
3
)
e
2
x
2
+
3
x
{\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=e^{2x^{2}+3x}(4x+3)=\mathbf {(4x+3)e^{2x^{2}+3x}} }
Let
f
(
x
)
=
g
(
h
(
x
)
)
;
g
(
x
)
=
e
x
;
h
(
x
)
=
2
x
2
+
3
x
{\displaystyle f(x)=g(h(x));\quad g(x)=e^{x};\quad h(x)=2x^{2}+3x}
. Then
f
′
(
x
)
=
d
g
d
h
d
h
d
x
=
e
2
x
2
+
3
x
(
4
x
+
3
)
=
(
4
x
+
3
)
e
2
x
2
+
3
x
{\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=e^{2x^{2}+3x}(4x+3)=\mathbf {(4x+3)e^{2x^{2}+3x}} }
56.
f
(
x
)
=
e
e
2
x
2
+
1
{\displaystyle f(x)=e^{e^{2x^{2}+1}}}
Logarithms
edit
59.
f
(
x
)
=
ln
(
ln
(
x
3
(
x
+
1
)
)
)
{\displaystyle f(x)=\ln(\ln(x^{3}(x+1)))\,}
Let
f
(
x
)
=
g
(
g
(
h
(
x
)
)
)
;
g
(
x
)
=
ln
(
x
)
;
h
(
x
)
=
x
3
(
x
+
1
)
{\displaystyle f(x)=g(g(h(x)));\quad g(x)=\ln(x);\quad h(x)=x^{3}(x+1)}
. Then
f
′
(
x
)
=
d
g
(
g
(
h
(
x
)
)
)
d
g
(
h
(
x
)
)
d
g
(
h
(
x
)
)
d
h
(
x
)
d
h
(
x
)
d
x
=
1
ln
(
x
3
(
x
+
1
)
)
1
x
3
(
x
+
1
)
(
4
x
3
+
3
x
2
)
=
4
x
3
+
3
x
2
x
3
(
x
+
1
)
ln
(
x
3
(
x
+
1
)
)
{\displaystyle f'(x)={\frac {dg(g(h(x)))}{dg(h(x))}}{\frac {dg(h(x))}{dh(x)}}{\frac {dh(x)}{dx}}={\frac {1}{\ln(x^{3}(x+1))}}{\frac {1}{x^{3}(x+1)}}(4x^{3}+3x^{2})=\mathbf {\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}} }
Let
f
(
x
)
=
g
(
g
(
h
(
x
)
)
)
;
g
(
x
)
=
ln
(
x
)
;
h
(
x
)
=
x
3
(
x
+
1
)
{\displaystyle f(x)=g(g(h(x)));\quad g(x)=\ln(x);\quad h(x)=x^{3}(x+1)}
. Then
f
′
(
x
)
=
d
g
(
g
(
h
(
x
)
)
)
d
g
(
h
(
x
)
)
d
g
(
h
(
x
)
)
d
h
(
x
)
d
h
(
x
)
d
x
=
1
ln
(
x
3
(
x
+
1
)
)
1
x
3
(
x
+
1
)
(
4
x
3
+
3
x
2
)
=
4
x
3
+
3
x
2
x
3
(
x
+
1
)
ln
(
x
3
(
x
+
1
)
)
{\displaystyle f'(x)={\frac {dg(g(h(x)))}{dg(h(x))}}{\frac {dg(h(x))}{dh(x)}}{\frac {dh(x)}{dx}}={\frac {1}{\ln(x^{3}(x+1))}}{\frac {1}{x^{3}(x+1)}}(4x^{3}+3x^{2})=\mathbf {\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}} }
Trigonometric functions
edit
More Differentiation
edit
67.
d
d
x
[
(
x
+
4
)
(
x
+
2
)
(
x
−
3
)
]
{\displaystyle {\frac {d}{dx}}[(x+4)(x+2)(x-3)]}
Let
f
(
x
)
=
A
(
x
)
B
(
x
)
C
(
x
)
;
A
(
x
)
=
x
+
4
;
B
(
x
)
=
x
+
2
;
C
(
x
)
=
x
−
3
{\displaystyle f(x)=A(x)B(x)C(x);\quad A(x)=x+4;\quad B(x)=x+2;\quad C(x)=x-3}
. Then
f
′
(
x
)
=
A
′
(
x
)
B
(
x
)
C
(
x
)
+
A
(
x
)
B
′
(
x
)
C
(
x
)
+
A
(
x
)
B
(
x
)
C
′
(
x
)
{\displaystyle f'(x)=A'(x)B(x)C(x)+A(x)B'(x)C(x)+A(x)B(x)C'(x)}
A
′
(
x
)
=
B
′
(
x
)
=
C
′
(
x
)
=
1
{\displaystyle A'(x)=B'(x)=C'(x)=1}
f
′
(
x
)
=
(
x
+
2
)
(
x
−
3
)
+
(
x
+
4
)
(
x
−
3
)
+
(
x
+
4
)
(
x
+
2
)
{\displaystyle f'(x)=\mathbf {(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)} }
Let
f
(
x
)
=
A
(
x
)
B
(
x
)
C
(
x
)
;
A
(
x
)
=
x
+
4
;
B
(
x
)
=
x
+
2
;
C
(
x
)
=
x
−
3
{\displaystyle f(x)=A(x)B(x)C(x);\quad A(x)=x+4;\quad B(x)=x+2;\quad C(x)=x-3}
. Then
f
′
(
x
)
=
A
′
(
x
)
B
(
x
)
C
(
x
)
+
A
(
x
)
B
′
(
x
)
C
(
x
)
+
A
(
x
)
B
(
x
)
C
′
(
x
)
{\displaystyle f'(x)=A'(x)B(x)C(x)+A(x)B'(x)C(x)+A(x)B(x)C'(x)}
A
′
(
x
)
=
B
′
(
x
)
=
C
′
(
x
)
=
1
{\displaystyle A'(x)=B'(x)=C'(x)=1}
f
′
(
x
)
=
(
x
+
2
)
(
x
−
3
)
+
(
x
+
4
)
(
x
−
3
)
+
(
x
+
4
)
(
x
+
2
)
{\displaystyle f'(x)=\mathbf {(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)} }
Implicit Differentiation
edit
Use implicit differentiation to find y'
74.
x
3
+
y
3
=
x
y
{\displaystyle x^{3}+y^{3}=xy\,}
75.
(
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\,}
Logarithmic Differentiation
edit
Use logarithmic differentiation to find
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
76.
y
=
x
(
1
−
x
3
4
)
{\displaystyle y=x({\sqrt[{4}]{1-x^{3}}}\,)}
77.
y
=
x
+
1
1
−
x
{\displaystyle y={\sqrt {x+1 \over 1-x}}\,}
78.
y
=
(
2
x
)
2
x
{\displaystyle y=(2x)^{2x}\,}
79.
y
=
(
x
3
+
4
x
)
3
x
+
1
{\displaystyle y=(x^{3}+4x)^{3x+1}\,}
80.
y
=
(
6
x
)
cos
(
x
)
+
1
{\displaystyle y=(6x)^{\cos(x)+1}\,}
Equation of Tangent Line
edit
For each function,
f
{\displaystyle f}
x
{\displaystyle x}
f
{\displaystyle f}
f
{\displaystyle f}
81.
f
(
x
)
=
x
3
3
+
x
2
+
5
,
(
3
,
23
)
{\displaystyle f(x)={\frac {x^{3}}{3}}+x^{2}+5,\;\;\;(3,23)}
82.
f
(
x
)
=
x
3
−
3
x
+
1
,
(
1
,
−
1
)
{\displaystyle f(x)=x^{3}-3x+1,\;\;\;(1,-1)}
83.
f
(
x
)
=
2
3
x
3
+
x
2
−
12
x
+
6
,
(
0
,
6
)
{\displaystyle f(x)={\frac {2}{3}}x^{3}+x^{2}-12x+6,\;\;\;(0,6)}
84.
f
(
x
)
=
2
x
+
1
x
,
(
1
,
3
)
{\displaystyle f(x)=2x+{\frac {1}{\sqrt {x}}},\;\;\;(1,3)}
85.
f
(
x
)
=
(
x
2
+
1
)
(
2
−
x
)
,
(
2
,
0
)
{\displaystyle f(x)=(x^{2}+1)(2-x),\;\;\;(2,0)}
86.
f
(
x
)
=
2
3
x
3
+
5
2
x
2
+
2
x
+
1
,
(
3
,
95
2
)
{\displaystyle f(x)={\frac {2}{3}}x^{3}+{\frac {5}{2}}x^{2}+2x+1,\;\;\;(3,{\frac {95}{2}})}
87. 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).
88. 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).
Higher Order Derivatives
edit
90. Use induction to prove that the (n+1)th derivative of a n-th order polynomial is 0.
Advanced Understanding of Derivatives
edit
91. Let
f
′
(
x
)
{\displaystyle f^{\prime }(x)}
be the derivative of
f
(
x
)
{\displaystyle f(x)}
. Prove the derivative of
−
f
(
x
)
{\displaystyle -f(x)}
is
−
f
′
(
x
)
{\displaystyle -f^{\prime }(x)}
.
Suppose
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
h
)
h
{\displaystyle f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}}
. Let
g
(
x
)
=
−
f
(
x
)
{\displaystyle g(x)=-f(x)}
.
g
′
(
x
)
=
lim
h
→
0
g
(
x
+
h
)
−
g
(
h
)
h
=
lim
h
→
0
−
f
(
x
+
h
)
+
f
(
h
)
h
=
lim
h
→
0
−
(
f
(
x
+
h
)
−
f
(
h
)
)
h
=
−
lim
h
→
0
(
f
(
x
+
h
)
−
f
(
h
)
)
h
=
−
f
′
(
x
)
{\displaystyle {\begin{aligned}g^{\prime }(x)&=\lim _{h\to 0}{\dfrac {g(x+h)-g(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-f(x+h)+f(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-\left(f(x+h)-f(h)\right)}{h}}\\&=-\lim _{h\to 0}{\dfrac {\left(f(x+h)-f(h)\right)}{h}}\\&=-f^{\prime }(x)\end{aligned}}}
Therefore, if
f
′
(
x
)
{\displaystyle f^{\prime }(x)}
is the derivative of
f
(
x
)
{\displaystyle f(x)}
, then
−
f
′
(
x
)
{\displaystyle -f^{\prime }(x)}
is the derivative of
−
f
(
x
)
{\displaystyle -f(x)}
.
◼
{\displaystyle \blacksquare }
Suppose
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
h
)
h
{\displaystyle f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}}
. Let
g
(
x
)
=
−
f
(
x
)
{\displaystyle g(x)=-f(x)}
.
g
′
(
x
)
=
lim
h
→
0
g
(
x
+
h
)
−
g
(
h
)
h
=
lim
h
→
0
−
f
(
x
+
h
)
+
f
(
h
)
h
=
lim
h
→
0
−
(
f
(
x
+
h
)
−
f
(
h
)
)
h
=
−
lim
h
→
0
(
f
(
x
+
h
)
−
f
(
h
)
)
h
=
−
f
′
(
x
)
{\displaystyle {\begin{aligned}g^{\prime }(x)&=\lim _{h\to 0}{\dfrac {g(x+h)-g(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-f(x+h)+f(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-\left(f(x+h)-f(h)\right)}{h}}\\&=-\lim _{h\to 0}{\dfrac {\left(f(x+h)-f(h)\right)}{h}}\\&=-f^{\prime }(x)\end{aligned}}}
Therefore, if
f
′
(
x
)
{\displaystyle f^{\prime }(x)}
is the derivative of
f
(
x
)
{\displaystyle f(x)}
, then
−
f
′
(
x
)
{\displaystyle -f^{\prime }(x)}
is the derivative of
−
f
(
x
)
{\displaystyle -f(x)}
.
◼
{\displaystyle \blacksquare }
92. Suppose a continuous function
f
(
x
)
{\displaystyle f(x)}
has three roots on the interval of
−
2
≤
x
≤
2
{\displaystyle -2\leq x\leq 2}
. If
f
(
−
2
)
=
f
(
2
)
<
0
{\displaystyle f(-2)=f(2)<0}
, then what is ONE true guarantee of
f
(
x
)
{\displaystyle f(x)}
using
(a) the Intermediate Value Theorem;
(b) Rolle's Theorem;
(c) the Extreme Value Theorem. 93. Let
g
(
x
)
=
f
−
1
(
x
)
{\displaystyle g(x)=f^{-1}(x)}
, where
f
−
1
(
x
)
{\displaystyle f^{-1}(x)}
is the inverse of
f
(
x
)
{\displaystyle f(x)}
. Let
f
(
x
)
{\displaystyle f(x)}
be differentiable. What is
g
′
(
x
)
{\displaystyle g^{\prime }(x)}
? Else, why can
g
′
(
x
)
{\displaystyle g^{\prime }(x)}
not be determined?
If
g
(
x
)
=
f
−
1
(
x
)
{\displaystyle g(x)=f^{-1}(x)}
, then
f
(
g
(
x
)
)
=
x
{\displaystyle f\left(g(x)\right)=x}
. We can use implicit differentiation.
d
d
x
[
f
(
g
(
x
)
)
]
=
d
d
x
(
x
)
g
′
(
x
)
f
′
(
g
(
x
)
)
=
1
g
′
(
x
)
=
1
f
′
(
g
(
x
)
)
=
1
f
′
(
f
−
1
(
x
)
)
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\left[f\left(g(x)\right)\right]&={\frac {d}{dx}}\left(x\right)\\g^{\prime }(x)f^{\prime }\left(g(x)\right)&=1\\g^{\prime }(x)&={\frac {1}{f^{\prime }\left(g(x)\right)}}\\&={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}\\\end{aligned}}}
If
g
(
x
)
=
f
−
1
(
x
)
{\displaystyle g(x)=f^{-1}(x)}
, then
f
(
g
(
x
)
)
=
x
{\displaystyle f\left(g(x)\right)=x}
. We can use implicit differentiation.
d
d
x
[
f
(
g
(
x
)
)
]
=
d
d
x
(
x
)
g
′
(
x
)
f
′
(
g
(
x
)
)
=
1
g
′
(
x
)
=
1
f
′
(
g
(
x
)
)
=
1
f
′
(
f
−
1
(
x
)
)
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\left[f\left(g(x)\right)\right]&={\frac {d}{dx}}\left(x\right)\\g^{\prime }(x)f^{\prime }\left(g(x)\right)&=1\\g^{\prime }(x)&={\frac {1}{f^{\prime }\left(g(x)\right)}}\\&={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}\\\end{aligned}}}
94. Let
f
(
x
)
=
{
2
x
2
+
4
,
x
<
−
1
a
x
3
−
2
x
,
x
≥
−
1
{\displaystyle f(x)={\begin{cases}2x^{2}+4,&x<-1\\ax^{3}-2x,&x\geq -1\end{cases}}}
where
a
{\displaystyle a}
is a constant.
Find a value, if possible, for
a
{\displaystyle a}
(a) The function
f
(
x
)
{\displaystyle f(x)}
(b) The function
f
(
x
)
{\displaystyle f(x)}
(a)
a
=
−
4
{\displaystyle a=-4}
.
2
(
−
1
)
2
+
4
=
a
(
−
1
)
3
−
2
(
−
1
)
⇒
a
=
−
4
{\displaystyle 2(-1)^{2}+4=a(-1)^{3}-2(-1)\Rightarrow a=-4}
f
′
(
x
)
=
{
4
x
,
x
<
−
1
3
a
x
2
−
2
,
x
≥
−
1
{\displaystyle f^{\prime }(x)={\begin{cases}4x,&x<-1\\3ax^{2}-2,&x\geq -1\end{cases}}}
4
(
−
1
)
≠
−
12
(
−
1
)
2
{\displaystyle 4(-1)\neq -12(-1)^{2}}
a
=
−
4
{\displaystyle a=-4}
(b) There is no
a
∈
R
{\displaystyle a\in \mathbb {R} }
A proof of this is simple.
lim
h
→
0
+
f
(
−
1
+
h
)
−
f
(
−
1
)
h
=
lim
h
→
0
+
2
(
−
1
+
h
)
2
+
4
−
4
−
2
h
=
lim
h
→
0
+
−
2
−
4
h
+
2
h
2
−
2
h
=
−
4
{\displaystyle {\begin{aligned}\lim _{h\to 0^{+}}{\frac {f(-1+h)-f(-1)}{h}}&=\lim _{h\to 0^{+}}{\frac {2(-1+h)^{2}+4-4-2}{h}}\\&=\lim _{h\to 0^{+}}{\frac {-2-4h+2h^{2}-2}{h}}\\&=-4\\\end{aligned}}}
However,
lim
h
→
0
−
f
(
−
1
+
h
)
−
f
(
−
1
)
h
=
lim
h
→
0
−
a
(
−
1
+
h
)
3
+
−
2
(
−
1
+
h
)
−
6
h
=
lim
h
→
0
−
a
(
1
+
3
h
−
3
h
2
+
h
3
)
+
2
−
2
h
−
6
h
=
lim
h
→
0
−
a
+
3
a
h
−
3
a
h
2
+
a
h
3
−
2
h
−
4
h
=
lim
h
→
0
−
3
a
h
−
3
a
h
2
+
a
h
3
−
2
h
+
(
a
−
4
)
h
{\displaystyle {\begin{aligned}\lim _{h\to 0^{-}}{\frac {f(-1+h)-f(-1)}{h}}&=\lim _{h\to 0^{-}}{\frac {a(-1+h)^{3}+-2(-1+h)-6}{h}}\\&=\lim _{h\to 0^{-}}{\frac {a\left(1+3h-3h^{2}+h^{3}\right)+2-2h-6}{h}}\\&=\lim _{h\to 0^{-}}{\frac {a+3ah-3ah^{2}+ah^{3}-2h-4}{h}}\\&=\lim _{h\to 0^{-}}{\frac {3ah-3ah^{2}+ah^{3}-2h+(a-4)}{h}}\end{aligned}}}
To allow the best possible chance, we will let
a
=
4
{\displaystyle a=4}
lim
h
→
0
−
3
(
4
)
h
−
3
(
4
)
h
2
+
(
4
)
h
3
−
2
h
+
(
4
−
4
)
h
=
lim
h
→
0
−
3
(
4
)
h
−
3
(
4
)
h
2
+
(
4
)
h
3
−
2
h
h
=
lim
h
→
0
−
12
−
12
h
+
4
h
2
−
2
=
10
{\displaystyle {\begin{aligned}\lim _{h\to 0^{-}}{\frac {3(4)h-3(4)h^{2}+(4)h^{3}-2h+(4-4)}{h}}&=\lim _{h\to 0^{-}}{\frac {3(4)h-3(4)h^{2}+(4)h^{3}-2h}{h}}\\&=\lim _{h\to 0^{-}}12-12h+4h^{2}-2\\&=10\end{aligned}}}
For any other
a
∈
R
{\displaystyle a\in \mathbb {R} }
a
∈
R
{\displaystyle a\in \mathbb {R} }
(a)
a
=
−
4
{\displaystyle a=-4}
.
2
(
−
1
)
2
+
4
=
a
(
−
1
)
3
−
2
(
−
1
)
⇒
a
=
−
4
{\displaystyle 2(-1)^{2}+4=a(-1)^{3}-2(-1)\Rightarrow a=-4}
f
′
(
x
)
=
{
4
x
,
x
<
−
1
3
a
x
2
−
2
,
x
≥
−
1
{\displaystyle f^{\prime }(x)={\begin{cases}4x,&x<-1\\3ax^{2}-2,&x\geq -1\end{cases}}}
4
(
−
1
)
≠
−
12
(
−
1
)
2
{\displaystyle 4(-1)\neq -12(-1)^{2}}
a
=
−
4
{\displaystyle a=-4}
(b) There is no
a
∈
R
{\displaystyle a\in \mathbb {R} }
A proof of this is simple.
lim
h
→
0
+
f
(
−
1
+
h
)
−
f
(
−
1
)
h
=
lim
h
→
0
+
2
(
−
1
+
h
)
2
+
4
−
4
−
2
h
=
lim
h
→
0
+
−
2
−
4
h
+
2
h
2
−
2
h
=
−
4
{\displaystyle {\begin{aligned}\lim _{h\to 0^{+}}{\frac {f(-1+h)-f(-1)}{h}}&=\lim _{h\to 0^{+}}{\frac {2(-1+h)^{2}+4-4-2}{h}}\\&=\lim _{h\to 0^{+}}{\frac {-2-4h+2h^{2}-2}{h}}\\&=-4\\\end{aligned}}}
However,
lim
h
→
0
−
f
(
−
1
+
h
)
−
f
(
−
1
)
h
=
lim
h
→
0
−
a
(
−
1
+
h
)
3
+
−
2
(
−
1
+
h
)
−
6
h
=
lim
h
→
0
−
a
(
1
+
3
h
−
3
h
2
+
h
3
)
+
2
−
2
h
−
6
h
=
lim
h
→
0
−
a
+
3
a
h
−
3
a
h
2
+
a
h
3
−
2
h
−
4
h
=
lim
h
→
0
−
3
a
h
−
3
a
h
2
+
a
h
3
−
2
h
+
(
a
−
4
)
h
{\displaystyle {\begin{aligned}\lim _{h\to 0^{-}}{\frac {f(-1+h)-f(-1)}{h}}&=\lim _{h\to 0^{-}}{\frac {a(-1+h)^{3}+-2(-1+h)-6}{h}}\\&=\lim _{h\to 0^{-}}{\frac {a\left(1+3h-3h^{2}+h^{3}\right)+2-2h-6}{h}}\\&=\lim _{h\to 0^{-}}{\frac {a+3ah-3ah^{2}+ah^{3}-2h-4}{h}}\\&=\lim _{h\to 0^{-}}{\frac {3ah-3ah^{2}+ah^{3}-2h+(a-4)}{h}}\end{aligned}}}
To allow the best possible chance, we will let
a
=
4
{\displaystyle a=4}
lim
h
→
0
−
3
(
4
)
h
−
3
(
4
)
h
2
+
(
4
)
h
3
−
2
h
+
(
4
−
4
)
h
=
lim
h
→
0
−
3
(
4
)
h
−
3
(
4
)
h
2
+
(
4
)
h
3
−
2
h
h
=
lim
h
→
0
−
12
−
12
h
+
4
h
2
−
2
=
10
{\displaystyle {\begin{aligned}\lim _{h\to 0^{-}}{\frac {3(4)h-3(4)h^{2}+(4)h^{3}-2h+(4-4)}{h}}&=\lim _{h\to 0^{-}}{\frac {3(4)h-3(4)h^{2}+(4)h^{3}-2h}{h}}\\&=\lim _{h\to 0^{-}}12-12h+4h^{2}-2\\&=10\end{aligned}}}
For any other
a
∈
R
{\displaystyle a\in \mathbb {R} }
a
∈
R
{\displaystyle a\in \mathbb {R} }
![{\displaystyle {\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {[2(x+\Delta x)+2]-(2x+2)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x+2\Delta x+2-2x-2}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x}}\\&=\mathbf {2} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2ff13db48115ad15ca0fb870ba1cfd847f64fd0)
![{\displaystyle {\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {[2(x+\Delta x)^{2}+4(x+\Delta x)+4]-(2x^{2}+4x+4)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {4x\Delta x+2\Delta x^{2}+4\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0}4x+2\Delta x+4\\&=\mathbf {4x+4} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/498835143e20132ee1839df6aaf684e54dfbb8b4)
![{\displaystyle {\frac {d}{dx}}\left[cf(x)\right]=c{\frac {d}{dx}}\left[f(x)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7257a7b32e3dd4d3ef6bfd2c6315f45c8cb0c561)
![{\displaystyle {\begin{aligned}{\frac {d}{dx}}\left[cf(x)\right]&=\lim _{\Delta x\to 0}{\frac {cf\left(x+\Delta x\right)-cf\left(x\right)}{\Delta x}}\\&=c\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\\&=c{\frac {d}{dx}}\left[f(x)\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcfc7ac4df3b8985d7d000ed9ca141d3f2b7a1d8)
![{\displaystyle f(x)=3{\sqrt[{3}]{x}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7b861dfb68b951e23d614557d2cb071669b0851)
![{\displaystyle f'(x)=3({\frac {1}{3}})x^{-2/3}=\mathbf {\frac {1}{\sqrt[{3}]{x^{2}}}} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/814676dda6261891b5e32ddebc8efb233ad4c411)
![{\displaystyle f'(x)={\frac {-2}{x^{3}}}+x^{-2/3}=\mathbf {{\frac {-2}{x^{3}}}+{\frac {1}{\sqrt[{3}]{x^{2}}}}} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a68756c6e8b616e4d0a37189a8058c881b7391c)
![{\displaystyle f(x)={\frac {3}{x^{4}}}-{\sqrt[{4}]{x}}+x\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e623fc9dc5921a33a5101f6a43b712a078ff7f12)
![{\displaystyle f'(x)={\frac {-12}{x^{5}}}-{\frac {1}{4}}x^{-3/4}+1=\mathbf {{\frac {-12}{x^{5}}}-{\frac {1}{4{\sqrt[{4}]{x^{3}}}}}+1} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/11ce424291dfead9cbff9b60892c61127980814d)
![{\displaystyle f'(x)=2x^{-2/3}-0.4x^{-0.6}-{\frac {18}{x^{3}}}=\mathbf {{\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/dd6cefc8f7b96dc0555a05043e35f044c8dc37ff)
![{\displaystyle f(x)={\frac {1}{\sqrt[{3}]{x}}}+{\sqrt {x}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6db0bee5880f21a927a7926cdb71d77a3a78e87)
![{\displaystyle f'(x)=-{\frac {1}{3x^{4/3}}}+{\frac {1}{2{\sqrt {x}}}}=\mathbf {{\frac {-1}{3x{\sqrt[{3}]{x}}}}+{\frac {1}{2{\sqrt {x}}}}} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e98e549cc15a160f8d8ee43f5d93c0795e27c668)
![{\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)
![{\displaystyle {\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/ba69464a1939efb019dcd0b74373bfa4681eb4f3)
![{\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/957f79e0769565723820a31ac23d16be33565100)
![{\displaystyle \ln y=\ln(x)+\ln({\sqrt[{4}]{1-x^{3}}})=\ln(x)+{\frac {\ln(1-x^{3})}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/195e7ddba61d3af67f37c87baa1ff4f6ab131d9e)
![{\displaystyle y'=x({\sqrt[{4}]{1-x^{3}}}\,)({\frac {1}{x}}-{\frac {3x^{2}}{4(1-x^{3})}})=\mathbf {{\sqrt[{4}]{1-x^{3}}}-{\frac {3x^{3}}{4(1-x^{3})^{3/4}}}} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c69b7531b08237887df2e00d200121bbd9e6539)
![{\displaystyle f(d)=c\in \left[f(-2),0\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e01c7ca0cd767dd136159e9462217fc21b9c2d7)
![{\displaystyle d\in \left[-2,2\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1c71d40549760ce0871c0d4f89deba85405eff4)
![{\displaystyle x\in \left[-2,2\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0628c1d0afabfc0fb43f3719fbf2eb8ccd4e516)
![{\displaystyle {\begin{aligned}{\frac {d}{dx}}\left[f\left(g(x)\right)\right]&={\frac {d}{dx}}\left(x\right)\\g^{\prime }(x)f^{\prime }\left(g(x)\right)&=1\\g^{\prime }(x)&={\frac {1}{f^{\prime }\left(g(x)\right)}}\\&={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5d8d63d67a2d7c246b198904ce6b548fbe8f4ea)
