SUCCESSIVE DIFFERENTIATION
74. Definition of successive derivatives. x is in general also a function of x . This new function may also be differentiable, in which case the derivative of the first derivative is called the second derivative of the original function. Similarly, the derivative of the second derivative is called the third derivative ; and so on to the n th derivative . Thus, if
y
{\displaystyle \ y}
=
3
x
4
{\displaystyle =\ 3x^{4}}
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
=
12
x
3
{\displaystyle =\ 12x^{3}}
d
d
x
(
d
y
d
x
)
{\displaystyle {\frac {d}{dx}}\left({\frac {dy}{dx}}\right)}
=
36
x
2
{\displaystyle =\ 36x^{2}}
d
d
x
[
d
d
x
(
d
y
d
x
)
]
{\displaystyle {\frac {d}{dx}}\left[{\frac {d}{dx}}\left({\frac {dy}{dx}}\right)\right]}
=
72
x
{\displaystyle =\ 72x}
75. Notation.
d
d
x
(
d
y
d
x
)
{\displaystyle {\frac {d}{dx}}\left({\frac {dy}{dx}}\right)}
=
d
2
y
d
x
2
{\displaystyle =\ {\frac {d^{2}y}{dx^{2}}}}
d
d
x
[
d
d
x
(
d
y
d
x
)
]
{\displaystyle {\frac {d}{dx}}\left[{\frac {d}{dx}}\left({\frac {dy}{dx}}\right)\right]}
=
d
d
x
(
d
2
y
d
x
2
)
=
d
3
y
d
x
3
{\displaystyle =\ {\frac {d}{dx}}\left({\frac {d^{2}y}{dx^{2}}}\right)={\frac {d^{3}y}{dx^{3}}}}
. . .
. . .
d
d
x
(
d
n
−
1
y
d
x
n
−
1
)
{\displaystyle {\frac {d}{dx}}\left({\frac {d^{n-1}y}{dx^{n-1}}}\right)}
=
d
n
y
d
x
n
{\displaystyle ={\frac {d^{n}y}{dx^{n}}}}
If
y
=
f
(
x
)
{\displaystyle y=f(x)}
f
′
(
x
)
,
f
″
(
x
)
,
f
‴
(
x
)
,
f
i
v
(
x
)
,
.
.
.
,
f
(
n
)
(
x
)
{\displaystyle f'(x),\ f''(x),\ f'''(x),\ f^{iv}(x),\ ...,\ f^{(n)}(x)}
y
′
,
y
″
,
y
‴
,
y
i
v
,
.
.
.
,
y
(
n
)
{\displaystyle y',\ y'',\ y''',\ y^{iv},\ ...,\ y^{(n)}}
or,
d
d
x
f
(
x
)
,
d
2
d
x
2
f
(
x
)
,
d
3
d
x
3
f
(
x
)
,
d
4
d
x
4
f
(
x
)
.
.
.
,
d
n
d
x
n
f
(
x
)
{\displaystyle {\frac {d}{dx}}f(x),\ {\frac {d^{2}}{dx^{2}}}f(x),\ {\frac {d^{3}}{dx^{3}}}f(x),\ {\frac {d^{4}}{dx^{4}}}f(x)\ ...,\ {\frac {d^{n}}{dx^{n}}}f(x)}
76. The n th derivative. n th derivative . The usual plan is to find a number of the first successive derivatives, as many as may be necessary to discover their law of formation, and then by induction write down the nth derivative.
Illustrative Example 1.
y
=
e
a
x
{\displaystyle y=e^{ax}}
d
n
y
d
x
n
{\displaystyle {\frac {d^{n}y}{dx^{n}}}}
Solution,
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
=
a
e
a
x
{\displaystyle =ae^{ax}}
d
2
y
d
x
2
{\displaystyle {\frac {d^{2}y}{dx^{2}}}}
=
a
2
e
a
x
{\displaystyle =a^{2}e^{ax}}
. . .
∴
d
n
y
d
x
n
{\displaystyle {\frac {d^{n}y}{dx^{n}}}}
=
a
n
e
a
x
{\displaystyle =a^{n}e^{ax}}
Ans.
Illustrative Example 2.
y
=
ln
x
{\displaystyle y=\ln x}
d
n
y
d
x
n
{\displaystyle {\frac {d^{n}y}{dx^{n}}}}
Solution,
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
=
1
x
{\displaystyle ={\frac {1}{x}}}
d
2
y
d
x
2
{\displaystyle {\frac {d^{2}y}{dx^{2}}}}
=
−
1
x
2
{\displaystyle =-{\frac {1}{x^{2}}}}
d
3
y
d
x
3
{\displaystyle {\frac {d^{3}y}{dx^{3}}}}
=
1
⋅
2
x
3
{\displaystyle ={\frac {1\cdot 2}{x^{3}}}}
d
4
y
d
x
4
{\displaystyle {\frac {d^{4}y}{dx^{4}}}}
=
1
⋅
2
⋅
3
x
4
{\displaystyle ={\frac {1\cdot 2\cdot 3}{x^{4}}}}
. . .
∴
d
n
y
d
x
n
{\displaystyle {\frac {d^{n}y}{dx^{n}}}}
=
(
−
1
)
n
−
1
(
n
−
1
)
!
x
n
{\displaystyle =(-1)^{n-1}{\frac {(n-1)!}{x^{n}}}}
Ans.
Illustrative Example 3.
y
=
sin
x
{\displaystyle y=\sin x}
d
n
y
d
x
n
{\displaystyle {\frac {d^{n}y}{dx^{n}}}}
Solution,
d
y
d
x
=
cos
x
{\displaystyle {\frac {dy}{dx}}=\cos x}
=
sin
(
x
+
π
2
)
{\displaystyle =\sin \left(x+{\frac {\pi }{2}}\right)}
d
2
y
d
x
2
=
d
d
x
sin
(
x
+
π
2
)
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\sin \left(x+{\frac {\pi }{2}}\right)}
=
cos
(
x
+
π
2
)
=
sin
(
x
+
2
π
2
)
{\displaystyle =\cos \left(x+{\frac {\pi }{2}}\right)=\sin \left(x+{\frac {2\pi }{2}}\right)}
d
3
y
d
x
3
=
d
d
x
sin
(
x
+
2
π
2
)
{\displaystyle {\frac {d^{3}y}{dx^{3}}}={\frac {d}{dx}}\sin \left(x+{\frac {2\pi }{2}}\right)}
=
cos
(
x
+
2
π
2
)
=
sin
(
x
+
3
π
2
)
{\displaystyle =\cos \left(x+{\frac {2\pi }{2}}\right)=\sin \left(x+{\frac {3\pi }{2}}\right)}
. . .
. . .
∴
d
n
y
d
x
n
{\displaystyle {\frac {d^{n}y}{dx^{n}}}}
=
sin
(
x
+
n
π
2
)
{\displaystyle =\sin \left(x+{\frac {n\pi }{2}}\right)}
Ans.
77. Leibnitz's Formula for the n th derivative of a product. n th derivative of the product of two variables in terms of the variables themselves and their successive derivatives.
If u and v are functions of x , we have, from V , §33
d
d
x
(
u
v
)
=
d
u
d
x
v
+
u
d
v
d
x
{\displaystyle {\frac {d}{dx}}(uv)={\frac {du}{dx}}v+u{\frac {dv}{dx}}}
Differentiating again with respect to x ,
d
2
d
x
2
(
u
v
)
=
d
2
u
d
x
2
v
+
d
u
d
x
d
v
d
x
+
d
u
d
x
d
v
d
x
+
u
d
2
v
d
x
2
=
d
2
u
d
x
2
v
+
2
d
u
d
x
d
v
d
x
+
u
d
2
v
d
x
2
{\displaystyle {\frac {d^{2}}{dx^{2}}}(uv)={\frac {d^{2}u}{dx^{2}}}v+{\frac {du}{dx}}{\frac {dv}{dx}}+{\frac {du}{dx}}{\frac {dv}{dx}}+u{\frac {d^{2}v}{dx^{2}}}={\frac {d^{2}u}{dx^{2}}}v+2{\frac {du}{dx}}{\frac {dv}{dx}}+u{\frac {d^{2}v}{dx^{2}}}}
Similarly,
d
3
d
x
3
(
u
v
)
{\displaystyle {\frac {d^{3}}{dx^{3}}}(uv)}
=
d
3
u
d
x
3
v
+
d
2
u
d
x
2
d
v
d
x
+
2
d
2
u
d
x
2
d
v
d
x
+
2
d
u
d
x
d
2
v
d
x
2
+
d
u
d
x
d
2
v
d
x
2
+
u
d
3
v
d
x
3
{\displaystyle ={\frac {d^{3}u}{dx^{3}}}v+{\frac {d^{2}u}{dx^{2}}}{\frac {dv}{dx}}+2{\frac {d^{2}u}{dx^{2}}}{\frac {dv}{dx}}+2{\frac {du}{dx}}{\frac {d^{2}v}{dx^{2}}}+{\frac {du}{dx}}{\frac {d^{2}v}{dx^{2}}}+u{\frac {d^{3}v}{dx^{3}}}}
=
d
3
u
d
x
3
v
+
3
d
2
u
d
x
2
d
v
d
x
+
3
d
u
d
x
d
2
v
d
x
2
+
u
d
3
v
d
x
3
{\displaystyle ={\frac {d^{3}u}{dx^{3}}}v+3{\frac {d^{2}u}{dx^{2}}}{\frac {dv}{dx}}+3{\frac {du}{dx}}{\frac {d^{2}v}{dx^{2}}}+u{\frac {d^{3}v}{dx^{3}}}}
However far this process may be continued, it will be seen that the numerical coefficients follow the same law as those of the Binomial Theorem, and the indices of the derivatives correspond to the exponents of the Binomial Theorem.[ 1] m th to the (m + 1 )th derivative of the product, we can prove Leibnitz's Formula
d
n
d
x
n
(
u
v
)
=
d
n
u
d
x
n
v
+
n
d
n
−
1
u
d
x
n
−
1
d
v
d
x
+
n
(
n
−
1
)
2
!
d
n
−
2
u
d
x
n
−
2
d
2
v
d
x
2
+
⋯
+
{\displaystyle {\frac {d^{n}}{dx^{n}}}(uv)={\frac {d^{n}u}{dx^{n}}}v+n{\frac {d^{n-1}u}{dx^{n-1}}}{\frac {dv}{dx}}+{\frac {n(n-1)}{2!}}{\frac {d^{n-2}u}{dx^{n-2}}}{\frac {d^{2}v}{dx^{2}}}+\cdots +}
n
d
u
d
x
d
n
−
1
v
d
x
n
−
1
+
u
d
n
d
x
n
v
{\displaystyle n{\frac {du}{dx}}{\frac {d^{n-1}v}{dx^{n-1}}}+u{\frac {d^{n}}{dx^{n}}}v}
Illustrative Example 1.
y
=
e
x
ln
x
{\displaystyle y=e^{x}\ln x}
d
3
y
d
x
3
{\displaystyle {\frac {d^{3}y}{dx^{3}}}}
Solution. Let
u
=
e
x
{\displaystyle u=e^{x}}
v
=
log
x
{\displaystyle v=\log x}
then
d
u
d
x
=
e
x
{\displaystyle {\frac {du}{dx}}=e^{x}}
d
v
d
x
=
1
x
{\displaystyle {\frac {dv}{dx}}={\frac {1}{x}}}
d
2
u
d
x
2
=
e
x
{\displaystyle {\frac {d^{2}u}{dx^{2}}}=e^{x}}
d
2
v
d
x
2
=
−
1
x
2
{\displaystyle {\frac {d^{2}v}{dx^{2}}}=-{\frac {1}{x^{2}}}}
d
3
u
d
x
3
=
e
x
{\displaystyle {\frac {d^{3}u}{dx^{3}}}=e^{x}}
d
3
v
d
x
3
=
2
x
3
{\displaystyle {\frac {d^{3}v}{dx^{3}}}={\frac {2}{x^{3}}}}
Substituting in (17 ) , we get
d
3
y
d
x
3
=
e
x
log
x
+
3
e
x
x
−
3
e
x
x
2
=
e
x
(
log
x
+
3
x
−
3
x
2
+
2
x
3
)
{\displaystyle {\frac {d^{3}y}{dx^{3}}}=e^{x}\log x+{\frac {3e^{x}}{x}}-{\frac {3e^{x}}{x^{2}}}=e^{x}\left(\log x+{\frac {3}{x}}-{\frac {3}{x^{2}}}+{\frac {2}{x^{3}}}\right)}
Illustrative Example 2.
y
=
x
2
e
a
x
{\displaystyle y=x^{2}e^{ax}}
d
n
y
d
x
n
{\displaystyle {\tfrac {d^{n}y}{dx^{n}}}}
Solution. Let
u
=
x
2
{\displaystyle u=x^{2}}
and
v
=
e
a
x
{\displaystyle v=e^{ax}}
then
d
u
d
x
=
2
x
{\displaystyle {\frac {du}{dx}}=2x}
d
v
d
x
=
a
e
a
x
{\displaystyle {\frac {dv}{dx}}=ae^{ax}}
d
2
u
d
x
2
=
2
{\displaystyle {\frac {d^{2}u}{dx^{2}}}=2}
d
2
v
d
x
2
=
a
2
e
a
x
{\displaystyle {\frac {d^{2}v}{dx^{2}}}=a^{2}e^{ax}}
d
3
u
d
x
3
=
0
{\displaystyle {\frac {d^{3}u}{dx^{3}}}=0}
d
3
v
d
x
3
=
a
3
e
a
x
{\displaystyle {\frac {d^{3}v}{dx^{3}}}=a^{3}e^{ax}}
. . .
. . .
d
n
u
d
x
n
=
0
{\displaystyle {\frac {d^{n}u}{dx^{n}}}=0}
d
n
v
d
x
n
=
a
n
e
a
x
{\displaystyle {\frac {d^{n}v}{dx^{n}}}=a^{n}e^{ax}}
Substituting in (17 ) , we get
d
n
y
d
x
n
=
x
2
a
n
e
a
x
+
2
n
a
n
−
1
x
e
a
x
+
n
(
n
−
1
)
a
n
−
2
e
a
x
=
a
n
−
2
e
a
x
[
x
2
a
2
+
2
n
a
x
+
n
(
n
−
1
)
]
{\displaystyle {\frac {d^{n}y}{dx^{n}}}=x^{2}a^{n}e^{ax}+2na^{n-1}xe^{ax}+n(n-1)a^{n-2}e^{ax}=a^{n-2}e^{ax}[x^{2}a^{2}+2nax+n(n-1)]}
78. Successive differentiation of implicit functions.
d
2
y
d
x
2
{\displaystyle {\frac {d^{2}y}{dx^{2}}}}
b
2
x
2
−
a
2
y
2
=
a
2
b
2
{\displaystyle b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2}}
Differentiating with respect to x , as in § 63 ,
2
b
2
x
−
2
a
2
y
d
y
d
x
=
0
{\displaystyle 2b^{2}x-2a^{2}y{\frac {dy}{dx}}=0}
or,
(A )
d
y
d
x
=
b
2
x
a
2
y
{\displaystyle {\frac {dy}{dx}}={\frac {b^{2}x}{a^{2}y}}}
Differentiating again, remembering that y is a function of x ,
d
2
y
d
x
2
=
a
2
y
b
2
−
b
2
x
a
2
d
y
d
x
a
4
y
2
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {a^{2}yb^{2}-b^{2}xa^{2}{\frac {dy}{dx}}}{a^{4}y^{2}}}}
Substituting for
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
(A ) ,
d
2
y
d
x
2
=
a
2
b
2
y
−
a
2
b
2
x
(
b
2
y
a
2
y
)
a
4
y
2
=
−
b
2
(
b
2
x
2
−
a
2
y
2
)
a
4
y
3
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {a^{2}b^{2}y-a^{2}b^{2}x\left({\frac {b^{2}y}{a^{2}y}}\right)}{a^{4}y^{2}}}=-{\frac {b^{2}(b^{2}x^{2}-a^{2}y^{2})}{a^{4}y^{3}}}}
But from the given equation,
b
2
x
2
−
a
2
y
2
=
a
2
b
2
{\displaystyle b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2}}
∴
d
2
y
d
x
2
=
−
b
4
a
2
y
3
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {b^{4}}{a^{2}y^{3}}}}
EXAMPLES
Differentiate the following:
1.
y
=
4
x
3
−
6
x
2
+
4
x
+
7
{\displaystyle y=4x^{3}-6x^{2}+4x+7}
d
2
y
d
x
2
=
12
(
2
x
−
1
)
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=12(2x-1)}
2.
f
(
x
)
=
x
3
1
−
x
{\displaystyle f(x)={\frac {x^{3}}{1-x}}}
f
i
v
(
x
)
=
4
!
(
1
−
x
)
5
{\displaystyle f^{iv}(x)={\frac {4!}{(1-x)^{5}}}}
3.
f
(
y
)
=
y
6
{\displaystyle f(y)=y^{6}}
f
v
i
(
y
)
=
6
!
{\displaystyle f^{vi}(y)=6!}
4.
y
=
x
3
log
x
{\displaystyle y=x^{3}\log x}
d
4
y
d
x
4
=
6
x
{\displaystyle {\frac {d^{4}y}{dx^{4}}}={\frac {6}{x}}}
5.
y
=
c
x
n
{\displaystyle y={\frac {c}{x^{n}}}}
y
″
=
n
(
n
+
1
)
c
x
n
+
2
{\displaystyle y''={\frac {n(n+1)c}{x^{n+2}}}}
6.
y
=
(
x
−
3
)
e
2
x
+
4
x
e
x
+
x
{\displaystyle y=(x-3)e^{2x}+4xe^{x}+x}
y
″
=
4
e
x
[
(
x
−
2
)
e
x
+
x
+
2
]
{\displaystyle y''=4e^{x}[(x-2)e^{x}+x+2]}
7.
y
=
a
2
(
e
x
a
+
e
−
x
a
)
{\displaystyle y={\frac {a}{2}}(e^{\frac {x}{a}}+e^{-{\frac {x}{a}}})}
y
″
=
1
2
a
(
e
x
a
+
e
−
x
a
)
=
y
a
2
{\displaystyle y''={\frac {1}{2a}}(e^{\frac {x}{a}}+e^{-{\frac {x}{a}}})={\frac {y}{a^{2}}}}
8.
f
(
x
)
=
a
x
2
+
b
x
+
c
{\displaystyle f(x)=ax^{2}+bx+c}
f
‴
(
x
)
=
0
{\displaystyle f'''(x)=0}
9.
f
(
x
)
=
log
(
x
+
1
)
{\displaystyle f(x)=\log(x+1)}
f
i
v
(
x
)
=
−
6
(
x
+
1
)
4
{\displaystyle f^{iv}(x)=-{\frac {6}{(x+1)^{4}}}}
10.
f
(
x
)
=
log
(
e
x
+
e
−
x
)
{\displaystyle f(x)=\log(e^{x}+e^{-x})}
f
‴
(
x
)
=
−
8
(
e
x
−
e
−
x
)
(
e
x
−
e
−
x
)
3
{\displaystyle f'''(x)=-{\frac {8(e^{x}-e^{-x})}{(e^{x}-e^{-x})^{3}}}}
11.
r
=
sin
a
θ
{\displaystyle r=\sin a\theta }
d
4
r
d
θ
4
=
a
4
sin
a
θ
=
a
4
r
{\displaystyle {\frac {d^{4}r}{d\theta ^{4}}}=a^{4}\sin a\theta =a^{4}r}
12.
r
=
tan
ϕ
{\displaystyle r=\tan \phi }
d
3
r
d
ϕ
3
=
6
sec
6
ϕ
−
4
sec
2
ϕ
{\displaystyle {\frac {d^{3}r}{d\phi ^{3}}}=6\sec ^{6}\phi -4\sec ^{2}\phi }
13.
r
=
log
sin
ϕ
{\displaystyle r=\log \sin \phi }
r
‴
=
2
cot
ϕ
csc
2
ϕ
{\displaystyle r'''=2\cot \phi \csc ^{2}\phi }
14.
f
(
t
)
=
e
−
t
cos
t
{\displaystyle f(t)=e^{-t}\cos t}
f
i
v
(
t
)
=
−
4
e
−
t
cos
t
=
−
4
f
(
t
)
{\displaystyle f^{iv}(t)=-4e^{-t}\cos t=-4f(t)}
15.
f
(
θ
)
=
sec
2
θ
{\displaystyle f(\theta )={\sqrt {\sec 2\theta }}}
f
″
(
θ
)
=
3
[
f
(
θ
)
]
5
−
f
(
θ
)
{\displaystyle f''(\theta )=3[f(\theta )]^{5}-f(\theta )}
16.
p
=
(
q
2
+
a
2
)
arctan
q
a
{\displaystyle p=(q^{2}+a^{2})\arctan {\frac {q}{a}}}
d
3
p
d
q
3
=
4
a
3
(
a
2
+
q
2
)
2
{\displaystyle {\frac {d^{3}p}{dq^{3}}}={\frac {4a^{3}}{(a^{2}+q^{2})^{2}}}}
17.
y
=
a
x
{\displaystyle y=a^{x}}
d
n
y
d
x
n
=
(
log
a
)
n
a
x
{\displaystyle {\frac {d^{n}y}{dx^{n}}}=(\log a)^{n}a^{x}}
18.
y
=
log
(
1
+
x
)
{\displaystyle y=\log(1+x)}
d
n
y
d
x
n
=
(
−
1
)
n
−
1
(
n
−
1
)
!
(
1
+
x
)
n
{\displaystyle {\frac {d^{n}y}{dx^{n}}}=(-1)^{n-1}{\frac {(n-1)!}{(1+x)^{n}}}}
19.
y
=
cos
a
x
{\displaystyle y=\cos ax}
d
n
y
d
x
n
=
a
n
cos
(
a
x
+
n
π
2
)
{\displaystyle {\frac {d^{n}y}{dx^{n}}}=a^{n}\cos \left(ax+{\frac {n\pi }{2}}\right)}
20.
y
=
x
n
−
1
log
x
{\displaystyle y=x^{n-1}\log x}
d
n
y
d
x
n
=
(
n
−
1
)
!
x
{\displaystyle {\frac {d^{n}y}{dx^{n}}}={\frac {(n-1)!}{x}}}
[n = a positive integer]
21.
y
=
1
−
x
1
+
x
{\displaystyle y={\frac {1-x}{1+x}}}
d
n
y
d
x
n
=
2
(
−
1
)
n
)
=
n
!
(
1
+
x
)
n
+
1
{\displaystyle {\frac {d^{n}y}{dx^{n}}}=2(-1)^{n})={\frac {n!}{(1+x)^{n+1}}}}
HINT. Reduce fraction to form
−
1
+
2
1
+
x
{\displaystyle -1+{\frac {2}{1+x}}}
22. If
y
=
e
x
sin
x
{\displaystyle y=e^{x}\sin x}
d
2
y
d
x
2
−
2
d
y
d
x
+
2
y
=
0
{\displaystyle {\frac {d^{2}y}{dx^{2}}}-2{\frac {dy}{dx}}+2y=0}
23. If
y
=
a
cos
(
log
x
)
+
b
sin
(
log
x
)
{\displaystyle y=a\cos(\log x)+b\sin(\log x)}
x
2
d
2
y
d
x
2
+
x
d
y
d
x
+
y
=
0
{\displaystyle x^{2}{\frac {d^{2}y}{dx^{2}}}+x{\frac {dy}{dx}}+y=0}
Use Leibnitz's Formula in the next four examples:
24.
y
=
x
2
a
x
{\displaystyle y=x^{2}a^{x}}
d
n
y
d
x
n
=
a
x
(
log
a
)
n
−
2
[
(
x
log
a
+
n
)
2
−
n
]
{\displaystyle {\frac {d^{n}y}{dx^{n}}}=a^{x}(\log a)^{n-2}[(x\log a+n)^{2}-n]}
25.
y
=
e
x
x
{\displaystyle y=e^{x}x}
d
n
y
d
x
n
=
e
x
(
x
+
n
)
{\displaystyle {\frac {d^{n}y}{dx^{n}}}=e^{x}(x+n)}
26.
f
(
x
)
=
e
x
sin
x
{\displaystyle f(x)=e^{x}\sin x}
f
(
n
)
(
x
)
=
(
2
)
n
e
x
s
i
n
(
x
+
n
π
4
)
{\displaystyle f^{(n)}(x)=({\sqrt {2}})^{n}e^{x}sin\left(x+{\frac {n\pi }{4}}\right)}
27.
f
(
θ
)
=
cos
a
θ
cos
b
θ
{\displaystyle f(\theta )=\cos a\theta \cos b\theta }
f
n
(
θ
)
=
(
a
+
b
)
n
2
cos
[
(
a
+
b
)
θ
+
n
π
2
]
{\displaystyle f^{n}(\theta )={\frac {(a+b)^{n}}{2}}\cos \left[(a+b)\theta +{\frac {n\pi }{2}}\right]}
+
(
a
−
b
)
n
2
cos
[
(
a
−
b
)
θ
+
n
π
2
]
{\displaystyle +{\frac {(a-b)^{n}}{2}}\cos \left[(a-b)\theta +{\frac {n\pi }{2}}\right]}
28. Show that the formulas for acceleration, (14 ) , (15 ) , §71 , may be written
α
=
d
2
s
d
t
2
,
α
x
=
d
2
x
d
t
2
,
α
y
=
d
2
y
d
t
2
{\displaystyle \alpha ={\frac {d^{2}s}{dt^{2}}},\alpha _{x}={\frac {d^{2}x}{dt^{2}}},\alpha _{y}={\frac {d^{2}y}{dt^{2}}}}
29.
y
2
=
4
a
x
{\displaystyle y^{2}=4ax}
d
2
y
d
x
2
=
−
4
a
2
y
3
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {4a^{2}}{y^{3}}}}
30.
b
2
x
2
+
a
2
Y
2
=
a
2
b
2
{\displaystyle b^{2}x^{2}+a^{2}Y^{2}=a^{2}b^{2}}
d
2
y
d
x
2
=
−
b
4
a
2
y
3
;
d
3
y
d
x
2
=
−
3
b
6
x
a
4
y
5
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {b^{4}}{a^{2}y^{3}}};{\frac {d^{3}y}{dx^{2}}}=-{\frac {3b^{6}x}{a^{4}y^{5}}}}
31.
x
2
+
y
2
=
r
2
{\displaystyle x^{2}+y^{2}=r^{2}}
d
2
y
d
x
2
=
−
r
2
y
3
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {r^{2}}{y^{3}}}}
32.
y
2
+
y
=
x
2
{\displaystyle y^{2}+y=x^{2}}
d
3
y
d
x
3
=
−
24
x
(
1
+
2
y
)
5
{\displaystyle {\frac {d^{3}y}{dx^{3}}}=-{\frac {24x}{(1+2y)^{5}}}}
33.
a
x
2
+
2
h
x
y
+
b
y
2
=
1
{\displaystyle ax^{2}+2hxy+by^{2}=1}
d
2
y
d
x
2
=
h
2
−
a
b
(
h
x
+
b
y
)
3
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {h^{2}-ab}{(hx+by)^{3}}}}
34.
y
2
−
2
x
y
=
a
2
{\displaystyle y^{2}-2xy=a^{2}}
d
2
y
d
x
2
=
a
2
(
y
−
x
)
3
;
d
3
y
d
x
3
=
−
3
a
2
x
(
y
−
x
)
5
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {a^{2}}{(y-x)^{3}}};{\frac {d^{3}y}{dx^{3}}}=-{\frac {3a^{2}x}{(y-x)^{5}}}}
35.
sec
ϕ
cos
θ
=
c
{\displaystyle \sec \phi \cos \theta =c}
d
2
θ
d
ϕ
2
=
tan
2
θ
−
tan
2
ϕ
tan
3
θ
{\displaystyle {\frac {d^{2}\theta }{d\phi ^{2}}}={\frac {\tan ^{2}\theta -\tan ^{2}\phi }{\tan ^{3}\theta }}}
36.
θ
tan
(
ϕ
+
θ
)
{\displaystyle \theta \tan(\phi +\theta )}
d
3
θ
d
ϕ
3
=
−
2
(
5
+
8
θ
2
+
3
θ
4
)
θ
8
{\displaystyle {\frac {d^{3}\theta }{d\phi ^{3}}}=-{\frac {2(5+8\theta ^{2}+3\theta ^{4})}{\theta ^{8}}}}
37. Find the second derivative in the following:
(a)
log
(
u
+
v
)
=
u
−
v
{\displaystyle \log(u+v)=u-v}
(e)
y
3
+
x
3
−
3
a
x
y
=
0
{\displaystyle y^{3}+x^{3}-3axy=0}
(b)
e
u
+
u
=
e
v
+
v
{\displaystyle e^{u}+u=e^{v}+v}
(f)
y
2
−
2
m
x
y
+
x
2
−
a
=
0
{\displaystyle y^{2}-2mxy+x^{2}-a=0}
(c)
s
=
1
+
t
e
s
{\displaystyle s=1+te^{s}}
(g)
y
=
sin
(
x
+
y
)
{\displaystyle y=\sin(x+y)}
(d)
e
s
+
s
t
−
e
=
0
{\displaystyle e^{s}+st-e=0}
(h)
e
x
+
y
=
x
y
{\displaystyle e^{x+y}=xy}
↑ To make this correspondence complete, u and v are considered as
d
0
u
d
x
0
{\displaystyle {\frac {d^{0}u}{dx^{0}}}}
d
0
v
d
d
x
0
{\displaystyle {\frac {d^{0}v}{ddx^{0}}}}
![{\displaystyle {\frac {d}{dx}}\left[{\frac {d}{dx}}\left({\frac {dy}{dx}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/584cd7fbc5c59c96cb3a031830ac2ef36b0bfc58)
![{\displaystyle {\frac {d^{n}y}{dx^{n}}}=x^{2}a^{n}e^{ax}+2na^{n-1}xe^{ax}+n(n-1)a^{n-2}e^{ax}=a^{n-2}e^{ax}[x^{2}a^{2}+2nax+n(n-1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6cc76d49bc7e6ef081498f642919519c6de94f6)
![{\displaystyle y''=4e^{x}[(x-2)e^{x}+x+2]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ce5ebf81b0ff63a734379b21205d7f016f4f5e1)
![{\displaystyle f''(\theta )=3[f(\theta )]^{5}-f(\theta )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66c96a1f518f5d39a2037fc5ee4b9e04b1171106)
![{\displaystyle {\frac {d^{n}y}{dx^{n}}}=a^{x}(\log a)^{n-2}[(x\log a+n)^{2}-n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c1554d081ea6f312d7d37414373d7ad16740a1d)
![{\displaystyle f^{n}(\theta )={\frac {(a+b)^{n}}{2}}\cos \left[(a+b)\theta +{\frac {n\pi }{2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c78cb460ca33441b6a9f9edccb7a3cd13f45a90)
![{\displaystyle +{\frac {(a-b)^{n}}{2}}\cos \left[(a-b)\theta +{\frac {n\pi }{2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13516bf024d9dc46eea3690bf948f7269db5a9d7)
