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1 | 2218-2221 | Two properties of ‘log’ functions are proved below:
(1)
There is a standard change of base rule to obtain loga p in terms of logb p Let
loga p = α, logb p = β and logb a = γ This means aα = p, bβ = p and bγ = a Substituting the third equation in the first one, we have
(bγ)α = bγα = p
Using this in the second equation, we get
bβ = p = bγα
which implies
β = αγ or α = β
γ |
1 | 2219-2222 | Let
loga p = α, logb p = β and logb a = γ This means aα = p, bβ = p and bγ = a Substituting the third equation in the first one, we have
(bγ)α = bγα = p
Using this in the second equation, we get
bβ = p = bγα
which implies
β = αγ or α = β
γ But then
loga p = log
log
b
b
p
a
(2)
Another interesting property of the log function is its effect on products |
1 | 2220-2223 | This means aα = p, bβ = p and bγ = a Substituting the third equation in the first one, we have
(bγ)α = bγα = p
Using this in the second equation, we get
bβ = p = bγα
which implies
β = αγ or α = β
γ But then
loga p = log
log
b
b
p
a
(2)
Another interesting property of the log function is its effect on products Let
logb pq = α |
1 | 2221-2224 | Substituting the third equation in the first one, we have
(bγ)α = bγα = p
Using this in the second equation, we get
bβ = p = bγα
which implies
β = αγ or α = β
γ But then
loga p = log
log
b
b
p
a
(2)
Another interesting property of the log function is its effect on products Let
logb pq = α Then bα = pq |
1 | 2222-2225 | But then
loga p = log
log
b
b
p
a
(2)
Another interesting property of the log function is its effect on products Let
logb pq = α Then bα = pq If logb p = β and logb q = γ, then bβ = p and bγ = q |
1 | 2223-2226 | Let
logb pq = α Then bα = pq If logb p = β and logb q = γ, then bβ = p and bγ = q But then bα = pq = bβbγ = bβ + γ
which implies α = β + γ, i |
1 | 2224-2227 | Then bα = pq If logb p = β and logb q = γ, then bβ = p and bγ = q But then bα = pq = bβbγ = bβ + γ
which implies α = β + γ, i e |
1 | 2225-2228 | If logb p = β and logb q = γ, then bβ = p and bγ = q But then bα = pq = bβbγ = bβ + γ
which implies α = β + γ, i e ,
logb pq = logb p + logb q
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
129
A particularly interesting and important consequence of this is when p = q |
1 | 2226-2229 | But then bα = pq = bβbγ = bβ + γ
which implies α = β + γ, i e ,
logb pq = logb p + logb q
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
129
A particularly interesting and important consequence of this is when p = q In
this case the above may be rewritten as
logb p2 = logb p + logb p = 2 log p
An easy generalisation of this (left as an exercise |
1 | 2227-2230 | e ,
logb pq = logb p + logb q
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
129
A particularly interesting and important consequence of this is when p = q In
this case the above may be rewritten as
logb p2 = logb p + logb p = 2 log p
An easy generalisation of this (left as an exercise ) is
logb pn = n log p
for any positive integer n |
1 | 2228-2231 | ,
logb pq = logb p + logb q
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
129
A particularly interesting and important consequence of this is when p = q In
this case the above may be rewritten as
logb p2 = logb p + logb p = 2 log p
An easy generalisation of this (left as an exercise ) is
logb pn = n log p
for any positive integer n In fact this is true for any real number n, but we will
not attempt to prove this |
1 | 2229-2232 | In
this case the above may be rewritten as
logb p2 = logb p + logb p = 2 log p
An easy generalisation of this (left as an exercise ) is
logb pn = n log p
for any positive integer n In fact this is true for any real number n, but we will
not attempt to prove this On the similar lines the reader is invited to verify
logb
x
y = logb x – logb y
Example 25 Is it true that x = elog x for all real x |
1 | 2230-2233 | ) is
logb pn = n log p
for any positive integer n In fact this is true for any real number n, but we will
not attempt to prove this On the similar lines the reader is invited to verify
logb
x
y = logb x – logb y
Example 25 Is it true that x = elog x for all real x Solution First, observe that the domain of log function is set of all positive real numbers |
1 | 2231-2234 | In fact this is true for any real number n, but we will
not attempt to prove this On the similar lines the reader is invited to verify
logb
x
y = logb x – logb y
Example 25 Is it true that x = elog x for all real x Solution First, observe that the domain of log function is set of all positive real numbers So the above equation is not true for non-positive real numbers |
1 | 2232-2235 | On the similar lines the reader is invited to verify
logb
x
y = logb x – logb y
Example 25 Is it true that x = elog x for all real x Solution First, observe that the domain of log function is set of all positive real numbers So the above equation is not true for non-positive real numbers Now, let y = elog x |
1 | 2233-2236 | Solution First, observe that the domain of log function is set of all positive real numbers So the above equation is not true for non-positive real numbers Now, let y = elog x If
y > 0, we may take logarithm which gives us log y = log (elog x) = log x |
1 | 2234-2237 | So the above equation is not true for non-positive real numbers Now, let y = elog x If
y > 0, we may take logarithm which gives us log y = log (elog x) = log x log e = log x |
1 | 2235-2238 | Now, let y = elog x If
y > 0, we may take logarithm which gives us log y = log (elog x) = log x log e = log x Thus
y = x |
1 | 2236-2239 | If
y > 0, we may take logarithm which gives us log y = log (elog x) = log x log e = log x Thus
y = x Hence x = elog x is true only for positive values of x |
1 | 2237-2240 | log e = log x Thus
y = x Hence x = elog x is true only for positive values of x One of the striking properties of the natural exponential function in differential
calculus is that it doesn’t change during the process of differentiation |
1 | 2238-2241 | Thus
y = x Hence x = elog x is true only for positive values of x One of the striking properties of the natural exponential function in differential
calculus is that it doesn’t change during the process of differentiation This is captured
in the following theorem whose proof we skip |
1 | 2239-2242 | Hence x = elog x is true only for positive values of x One of the striking properties of the natural exponential function in differential
calculus is that it doesn’t change during the process of differentiation This is captured
in the following theorem whose proof we skip Theorem 5*
(1)
The derivative of ex w |
1 | 2240-2243 | One of the striking properties of the natural exponential function in differential
calculus is that it doesn’t change during the process of differentiation This is captured
in the following theorem whose proof we skip Theorem 5*
(1)
The derivative of ex w r |
1 | 2241-2244 | This is captured
in the following theorem whose proof we skip Theorem 5*
(1)
The derivative of ex w r t |
1 | 2242-2245 | Theorem 5*
(1)
The derivative of ex w r t , x is ex; i |
1 | 2243-2246 | r t , x is ex; i e |
1 | 2244-2247 | t , x is ex; i e , d
dx (ex) = ex |
1 | 2245-2248 | , x is ex; i e , d
dx (ex) = ex (2)
The derivative of log x w |
1 | 2246-2249 | e , d
dx (ex) = ex (2)
The derivative of log x w r |
1 | 2247-2250 | , d
dx (ex) = ex (2)
The derivative of log x w r t |
1 | 2248-2251 | (2)
The derivative of log x w r t , x is 1
x ; i |
1 | 2249-2252 | r t , x is 1
x ; i e |
1 | 2250-2253 | t , x is 1
x ; i e , d
dx (log x) = 1
x |
1 | 2251-2254 | , x is 1
x ; i e , d
dx (log x) = 1
x Example 26 Differentiate the following w |
1 | 2252-2255 | e , d
dx (log x) = 1
x Example 26 Differentiate the following w r |
1 | 2253-2256 | , d
dx (log x) = 1
x Example 26 Differentiate the following w r t |
1 | 2254-2257 | Example 26 Differentiate the following w r t x:
(i)
e–x
(ii) sin (log x), x > 0
(iii) cos–1 (ex)
(iv) ecos x
Solution
(i)
Let y = e– x |
1 | 2255-2258 | r t x:
(i)
e–x
(ii) sin (log x), x > 0
(iii) cos–1 (ex)
(iv) ecos x
Solution
(i)
Let y = e– x Using chain rule, we have
dy
dx =
x
d
e
dx
− ⋅
(– x) = – e– x
(ii)
Let y = sin (log x) |
1 | 2256-2259 | t x:
(i)
e–x
(ii) sin (log x), x > 0
(iii) cos–1 (ex)
(iv) ecos x
Solution
(i)
Let y = e– x Using chain rule, we have
dy
dx =
x
d
e
dx
− ⋅
(– x) = – e– x
(ii)
Let y = sin (log x) Using chain rule, we have
dy
dx =
cos (log )
cos (log )
d(log )
x
x
x
dx
x
⋅
=
* Please see supplementary material on Page 222 |
1 | 2257-2260 | x:
(i)
e–x
(ii) sin (log x), x > 0
(iii) cos–1 (ex)
(iv) ecos x
Solution
(i)
Let y = e– x Using chain rule, we have
dy
dx =
x
d
e
dx
− ⋅
(– x) = – e– x
(ii)
Let y = sin (log x) Using chain rule, we have
dy
dx =
cos (log )
cos (log )
d(log )
x
x
x
dx
x
⋅
=
* Please see supplementary material on Page 222 Rationalised 2023-24
MATHEMATICS
130
(iii)
Let y = cos–1 (ex) |
1 | 2258-2261 | Using chain rule, we have
dy
dx =
x
d
e
dx
− ⋅
(– x) = – e– x
(ii)
Let y = sin (log x) Using chain rule, we have
dy
dx =
cos (log )
cos (log )
d(log )
x
x
x
dx
x
⋅
=
* Please see supplementary material on Page 222 Rationalised 2023-24
MATHEMATICS
130
(iii)
Let y = cos–1 (ex) Using chain rule, we have
dy
dx =
2
2
1
(
)
1
(
)
1
x
x
x
x
d
e
dxe
e
e
−
−
⋅
=
−
−
(iv)
Let y = ecos x |
1 | 2259-2262 | Using chain rule, we have
dy
dx =
cos (log )
cos (log )
d(log )
x
x
x
dx
x
⋅
=
* Please see supplementary material on Page 222 Rationalised 2023-24
MATHEMATICS
130
(iii)
Let y = cos–1 (ex) Using chain rule, we have
dy
dx =
2
2
1
(
)
1
(
)
1
x
x
x
x
d
e
dxe
e
e
−
−
⋅
=
−
−
(iv)
Let y = ecos x Using chain rule, we have
dy
dx =
cos
cos
( sin )
(sin )
x
x
e
x
x e
⋅ −
= −
EXERCISE 5 |
1 | 2260-2263 | Rationalised 2023-24
MATHEMATICS
130
(iii)
Let y = cos–1 (ex) Using chain rule, we have
dy
dx =
2
2
1
(
)
1
(
)
1
x
x
x
x
d
e
dxe
e
e
−
−
⋅
=
−
−
(iv)
Let y = ecos x Using chain rule, we have
dy
dx =
cos
cos
( sin )
(sin )
x
x
e
x
x e
⋅ −
= −
EXERCISE 5 4
Differentiate the following w |
1 | 2261-2264 | Using chain rule, we have
dy
dx =
2
2
1
(
)
1
(
)
1
x
x
x
x
d
e
dxe
e
e
−
−
⋅
=
−
−
(iv)
Let y = ecos x Using chain rule, we have
dy
dx =
cos
cos
( sin )
(sin )
x
x
e
x
x e
⋅ −
= −
EXERCISE 5 4
Differentiate the following w r |
1 | 2262-2265 | Using chain rule, we have
dy
dx =
cos
cos
( sin )
(sin )
x
x
e
x
x e
⋅ −
= −
EXERCISE 5 4
Differentiate the following w r t |
1 | 2263-2266 | 4
Differentiate the following w r t x:
1 |
1 | 2264-2267 | r t x:
1 sin
xe
x
2 |
1 | 2265-2268 | t x:
1 sin
xe
x
2 sin1
x
e
−
3 |
1 | 2266-2269 | x:
1 sin
xe
x
2 sin1
x
e
−
3 3xe
4 |
1 | 2267-2270 | sin
xe
x
2 sin1
x
e
−
3 3xe
4 sin (tan–1 e–x)
5 |
1 | 2268-2271 | sin1
x
e
−
3 3xe
4 sin (tan–1 e–x)
5 log (cos ex)
6 |
1 | 2269-2272 | 3xe
4 sin (tan–1 e–x)
5 log (cos ex)
6 2
5 |
1 | 2270-2273 | sin (tan–1 e–x)
5 log (cos ex)
6 2
5 x
x
x
e
e
e
+
+
+
7 |
1 | 2271-2274 | log (cos ex)
6 2
5 x
x
x
e
e
e
+
+
+
7 ,
0
ex
x >
8 |
1 | 2272-2275 | 2
5 x
x
x
e
e
e
+
+
+
7 ,
0
ex
x >
8 log (log x), x > 1
9 |
1 | 2273-2276 | x
x
x
e
e
e
+
+
+
7 ,
0
ex
x >
8 log (log x), x > 1
9 cos ,
0
log
x
x
x
>
10 |
1 | 2274-2277 | ,
0
ex
x >
8 log (log x), x > 1
9 cos ,
0
log
x
x
x
>
10 cos (log x + ex), x > 0
5 |
1 | 2275-2278 | log (log x), x > 1
9 cos ,
0
log
x
x
x
>
10 cos (log x + ex), x > 0
5 5 |
1 | 2276-2279 | cos ,
0
log
x
x
x
>
10 cos (log x + ex), x > 0
5 5 Logarithmic Differentiation
In this section, we will learn to differentiate certain special class of functions given in
the form
y = f (x) = [u(x)]v (x)
By taking logarithm (to base e) the above may be rewritten as
log y = v(x) log [u(x)]
Using chain rule we may differentiate this to get
1
1
( )
( )
dy
v x
y dx
u x
⋅
=
⋅ |
1 | 2277-2280 | cos (log x + ex), x > 0
5 5 Logarithmic Differentiation
In this section, we will learn to differentiate certain special class of functions given in
the form
y = f (x) = [u(x)]v (x)
By taking logarithm (to base e) the above may be rewritten as
log y = v(x) log [u(x)]
Using chain rule we may differentiate this to get
1
1
( )
( )
dy
v x
y dx
u x
⋅
=
⋅ u′(x) + v′(x) |
1 | 2278-2281 | 5 Logarithmic Differentiation
In this section, we will learn to differentiate certain special class of functions given in
the form
y = f (x) = [u(x)]v (x)
By taking logarithm (to base e) the above may be rewritten as
log y = v(x) log [u(x)]
Using chain rule we may differentiate this to get
1
1
( )
( )
dy
v x
y dx
u x
⋅
=
⋅ u′(x) + v′(x) log [u(x)]
which implies that
[
]
( )
( )
( ) log
( )
( )
dy
yv x
u x
v x
u x
dx
u x
=
⋅ ′
+ ′
⋅
The main point to be noted in this method is that f(x) and u(x) must always be
positive as otherwise their logarithms are not defined |
1 | 2279-2282 | Logarithmic Differentiation
In this section, we will learn to differentiate certain special class of functions given in
the form
y = f (x) = [u(x)]v (x)
By taking logarithm (to base e) the above may be rewritten as
log y = v(x) log [u(x)]
Using chain rule we may differentiate this to get
1
1
( )
( )
dy
v x
y dx
u x
⋅
=
⋅ u′(x) + v′(x) log [u(x)]
which implies that
[
]
( )
( )
( ) log
( )
( )
dy
yv x
u x
v x
u x
dx
u x
=
⋅ ′
+ ′
⋅
The main point to be noted in this method is that f(x) and u(x) must always be
positive as otherwise their logarithms are not defined This process of differentiation is
known as logarithms differentiation and is illustrated by the following examples:
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
131
Example 27 Differentiate
2
2
(
3) (
4)
3
4
5
x
x
x
x
−
+
+
+
w |
1 | 2280-2283 | u′(x) + v′(x) log [u(x)]
which implies that
[
]
( )
( )
( ) log
( )
( )
dy
yv x
u x
v x
u x
dx
u x
=
⋅ ′
+ ′
⋅
The main point to be noted in this method is that f(x) and u(x) must always be
positive as otherwise their logarithms are not defined This process of differentiation is
known as logarithms differentiation and is illustrated by the following examples:
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
131
Example 27 Differentiate
2
2
(
3) (
4)
3
4
5
x
x
x
x
−
+
+
+
w r |
1 | 2281-2284 | log [u(x)]
which implies that
[
]
( )
( )
( ) log
( )
( )
dy
yv x
u x
v x
u x
dx
u x
=
⋅ ′
+ ′
⋅
The main point to be noted in this method is that f(x) and u(x) must always be
positive as otherwise their logarithms are not defined This process of differentiation is
known as logarithms differentiation and is illustrated by the following examples:
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
131
Example 27 Differentiate
2
2
(
3) (
4)
3
4
5
x
x
x
x
−
+
+
+
w r t |
1 | 2282-2285 | This process of differentiation is
known as logarithms differentiation and is illustrated by the following examples:
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
131
Example 27 Differentiate
2
2
(
3) (
4)
3
4
5
x
x
x
x
−
+
+
+
w r t x |
1 | 2283-2286 | r t x Solution Let
2
2
(
3) (
4)
(3
4
5)
x
x
y
x
x
−
+
=
+
+
Taking logarithm on both sides, we have
log y = 1
2 [log (x – 3) + log (x2 + 4) – log (3x2 + 4x + 5)]
Now, differentiating both sides w |
1 | 2284-2287 | t x Solution Let
2
2
(
3) (
4)
(3
4
5)
x
x
y
x
x
−
+
=
+
+
Taking logarithm on both sides, we have
log y = 1
2 [log (x – 3) + log (x2 + 4) – log (3x2 + 4x + 5)]
Now, differentiating both sides w r |
1 | 2285-2288 | x Solution Let
2
2
(
3) (
4)
(3
4
5)
x
x
y
x
x
−
+
=
+
+
Taking logarithm on both sides, we have
log y = 1
2 [log (x – 3) + log (x2 + 4) – log (3x2 + 4x + 5)]
Now, differentiating both sides w r t |
1 | 2286-2289 | Solution Let
2
2
(
3) (
4)
(3
4
5)
x
x
y
x
x
−
+
=
+
+
Taking logarithm on both sides, we have
log y = 1
2 [log (x – 3) + log (x2 + 4) – log (3x2 + 4x + 5)]
Now, differentiating both sides w r t x, we get
1 dy
y dx
⋅
=
2
2
1
1
2
6
4
2 (
3)
4
3
4
5
x
x
x
x
x
+x
+
−
−
+
+
+
or
dy
dx =
2
2
1
2
6
4
2
(
3)
4
3
4
5
y
x
x
x
x
x
+x
+
−
−
+
+
+
=
2
2
2
2
1
(
3)(
4)
1
2
6
4
2
(
3)
3
4
5
4
3
4
5
x
x
x
x
x
x
x
x
x
x
−
+
+
+
−
−
+
+
+
+
+
Example 28 Differentiate ax w |
1 | 2287-2290 | r t x, we get
1 dy
y dx
⋅
=
2
2
1
1
2
6
4
2 (
3)
4
3
4
5
x
x
x
x
x
+x
+
−
−
+
+
+
or
dy
dx =
2
2
1
2
6
4
2
(
3)
4
3
4
5
y
x
x
x
x
x
+x
+
−
−
+
+
+
=
2
2
2
2
1
(
3)(
4)
1
2
6
4
2
(
3)
3
4
5
4
3
4
5
x
x
x
x
x
x
x
x
x
x
−
+
+
+
−
−
+
+
+
+
+
Example 28 Differentiate ax w r |
1 | 2288-2291 | t x, we get
1 dy
y dx
⋅
=
2
2
1
1
2
6
4
2 (
3)
4
3
4
5
x
x
x
x
x
+x
+
−
−
+
+
+
or
dy
dx =
2
2
1
2
6
4
2
(
3)
4
3
4
5
y
x
x
x
x
x
+x
+
−
−
+
+
+
=
2
2
2
2
1
(
3)(
4)
1
2
6
4
2
(
3)
3
4
5
4
3
4
5
x
x
x
x
x
x
x
x
x
x
−
+
+
+
−
−
+
+
+
+
+
Example 28 Differentiate ax w r t |
1 | 2289-2292 | x, we get
1 dy
y dx
⋅
=
2
2
1
1
2
6
4
2 (
3)
4
3
4
5
x
x
x
x
x
+x
+
−
−
+
+
+
or
dy
dx =
2
2
1
2
6
4
2
(
3)
4
3
4
5
y
x
x
x
x
x
+x
+
−
−
+
+
+
=
2
2
2
2
1
(
3)(
4)
1
2
6
4
2
(
3)
3
4
5
4
3
4
5
x
x
x
x
x
x
x
x
x
x
−
+
+
+
−
−
+
+
+
+
+
Example 28 Differentiate ax w r t x, where a is a positive constant |
1 | 2290-2293 | r t x, where a is a positive constant Solution Let y = ax |
1 | 2291-2294 | t x, where a is a positive constant Solution Let y = ax Then
log y = x log a
Differentiating both sides w |
1 | 2292-2295 | x, where a is a positive constant Solution Let y = ax Then
log y = x log a
Differentiating both sides w r |
1 | 2293-2296 | Solution Let y = ax Then
log y = x log a
Differentiating both sides w r t |
1 | 2294-2297 | Then
log y = x log a
Differentiating both sides w r t x, we have
1 dy
y dx = log a
or
dy
dx = y log a
Thus
(
x)
d
dxa
= ax log a
Alternatively
(
x)
d
dxa
=
log
log
(
)
( log )
x
a
x
a
d
d
e
e
x
a
dx
dx
=
= ex log a |
1 | 2295-2298 | r t x, we have
1 dy
y dx = log a
or
dy
dx = y log a
Thus
(
x)
d
dxa
= ax log a
Alternatively
(
x)
d
dxa
=
log
log
(
)
( log )
x
a
x
a
d
d
e
e
x
a
dx
dx
=
= ex log a log a = ax log a |
1 | 2296-2299 | t x, we have
1 dy
y dx = log a
or
dy
dx = y log a
Thus
(
x)
d
dxa
= ax log a
Alternatively
(
x)
d
dxa
=
log
log
(
)
( log )
x
a
x
a
d
d
e
e
x
a
dx
dx
=
= ex log a log a = ax log a Rationalised 2023-24
MATHEMATICS
132
Example 29 Differentiate xsin x, x > 0 w |
1 | 2297-2300 | x, we have
1 dy
y dx = log a
or
dy
dx = y log a
Thus
(
x)
d
dxa
= ax log a
Alternatively
(
x)
d
dxa
=
log
log
(
)
( log )
x
a
x
a
d
d
e
e
x
a
dx
dx
=
= ex log a log a = ax log a Rationalised 2023-24
MATHEMATICS
132
Example 29 Differentiate xsin x, x > 0 w r |
1 | 2298-2301 | log a = ax log a Rationalised 2023-24
MATHEMATICS
132
Example 29 Differentiate xsin x, x > 0 w r t |
1 | 2299-2302 | Rationalised 2023-24
MATHEMATICS
132
Example 29 Differentiate xsin x, x > 0 w r t x |
1 | 2300-2303 | r t x Solution Let y = xsin x |
1 | 2301-2304 | t x Solution Let y = xsin x Taking logarithm on both sides, we have
log y = sin x log x
Therefore
1 |
1 | 2302-2305 | x Solution Let y = xsin x Taking logarithm on both sides, we have
log y = sin x log x
Therefore
1 dy
y dx = sin
(log )
log
(sin )
d
d
x
x
x
x
dx
dx
+
or
1 dy
y dx =
(sin )1
log
cos
x
x
x
x +
or
dy
dx =
sin
cos
log
x
y
x
x
x
+
=
sin
sin
cos
log
x
x
x
x
x
x
+
=
sin
1
sin
sin
cos
log
x
x
x
x
x
x
x
− ⋅
+
⋅
Example 30 Find dy
dx , if yx + xy + xx = ab |
1 | 2303-2306 | Solution Let y = xsin x Taking logarithm on both sides, we have
log y = sin x log x
Therefore
1 dy
y dx = sin
(log )
log
(sin )
d
d
x
x
x
x
dx
dx
+
or
1 dy
y dx =
(sin )1
log
cos
x
x
x
x +
or
dy
dx =
sin
cos
log
x
y
x
x
x
+
=
sin
sin
cos
log
x
x
x
x
x
x
+
=
sin
1
sin
sin
cos
log
x
x
x
x
x
x
x
− ⋅
+
⋅
Example 30 Find dy
dx , if yx + xy + xx = ab Solution Given that yx + xy + xx = ab |
1 | 2304-2307 | Taking logarithm on both sides, we have
log y = sin x log x
Therefore
1 dy
y dx = sin
(log )
log
(sin )
d
d
x
x
x
x
dx
dx
+
or
1 dy
y dx =
(sin )1
log
cos
x
x
x
x +
or
dy
dx =
sin
cos
log
x
y
x
x
x
+
=
sin
sin
cos
log
x
x
x
x
x
x
+
=
sin
1
sin
sin
cos
log
x
x
x
x
x
x
x
− ⋅
+
⋅
Example 30 Find dy
dx , if yx + xy + xx = ab Solution Given that yx + xy + xx = ab Putting u = yx, v = xy and w = xx, we get u + v + w = ab
Therefore
0
du
dv
dw
dx
dx
dx
+
+
= |
1 | 2305-2308 | dy
y dx = sin
(log )
log
(sin )
d
d
x
x
x
x
dx
dx
+
or
1 dy
y dx =
(sin )1
log
cos
x
x
x
x +
or
dy
dx =
sin
cos
log
x
y
x
x
x
+
=
sin
sin
cos
log
x
x
x
x
x
x
+
=
sin
1
sin
sin
cos
log
x
x
x
x
x
x
x
− ⋅
+
⋅
Example 30 Find dy
dx , if yx + xy + xx = ab Solution Given that yx + xy + xx = ab Putting u = yx, v = xy and w = xx, we get u + v + w = ab
Therefore
0
du
dv
dw
dx
dx
dx
+
+
= (1)
Now, u = yx |
1 | 2306-2309 | Solution Given that yx + xy + xx = ab Putting u = yx, v = xy and w = xx, we get u + v + w = ab
Therefore
0
du
dv
dw
dx
dx
dx
+
+
= (1)
Now, u = yx Taking logarithm on both sides, we have
log u = x log y
Differentiating both sides w |
1 | 2307-2310 | Putting u = yx, v = xy and w = xx, we get u + v + w = ab
Therefore
0
du
dv
dw
dx
dx
dx
+
+
= (1)
Now, u = yx Taking logarithm on both sides, we have
log u = x log y
Differentiating both sides w r |
1 | 2308-2311 | (1)
Now, u = yx Taking logarithm on both sides, we have
log u = x log y
Differentiating both sides w r t |
1 | 2309-2312 | Taking logarithm on both sides, we have
log u = x log y
Differentiating both sides w r t x, we have
1 du
u dx
⋅
=
(log )
log
( )
d
d
x
y
y
x
dx
dx
+
=
1
log
1
dy
x
y
y dx
⋅
+
⋅
So
du
dx =
log
log
x
x dy
x dy
u
y
y
y
y dx
y dx
+
=
+
|
1 | 2310-2313 | r t x, we have
1 du
u dx
⋅
=
(log )
log
( )
d
d
x
y
y
x
dx
dx
+
=
1
log
1
dy
x
y
y dx
⋅
+
⋅
So
du
dx =
log
log
x
x dy
x dy
u
y
y
y
y dx
y dx
+
=
+
(2)
Also v = xy
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
133
Taking logarithm on both sides, we have
log v = y log x
Differentiating both sides w |
1 | 2311-2314 | t x, we have
1 du
u dx
⋅
=
(log )
log
( )
d
d
x
y
y
x
dx
dx
+
=
1
log
1
dy
x
y
y dx
⋅
+
⋅
So
du
dx =
log
log
x
x dy
x dy
u
y
y
y
y dx
y dx
+
=
+
(2)
Also v = xy
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
133
Taking logarithm on both sides, we have
log v = y log x
Differentiating both sides w r |
1 | 2312-2315 | x, we have
1 du
u dx
⋅
=
(log )
log
( )
d
d
x
y
y
x
dx
dx
+
=
1
log
1
dy
x
y
y dx
⋅
+
⋅
So
du
dx =
log
log
x
x dy
x dy
u
y
y
y
y dx
y dx
+
=
+
(2)
Also v = xy
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
133
Taking logarithm on both sides, we have
log v = y log x
Differentiating both sides w r t |
1 | 2313-2316 | (2)
Also v = xy
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
133
Taking logarithm on both sides, we have
log v = y log x
Differentiating both sides w r t x, we have
1 dv
v dx
⋅
=
(log )
log
d
dy
y
x
x
dx
dx
+
=
1
log
dy
y
x
x
dx
⋅
+
⋅
So
dv
dx =
log
y
dy
v
x
x
dx
+
=
log
y
y
dy
x
x
x
dx
+
|
1 | 2314-2317 | r t x, we have
1 dv
v dx
⋅
=
(log )
log
d
dy
y
x
x
dx
dx
+
=
1
log
dy
y
x
x
dx
⋅
+
⋅
So
dv
dx =
log
y
dy
v
x
x
dx
+
=
log
y
y
dy
x
x
x
dx
+
(3)
Again
w = xx
Taking logarithm on both sides, we have
log w = x log x |
1 | 2315-2318 | t x, we have
1 dv
v dx
⋅
=
(log )
log
d
dy
y
x
x
dx
dx
+
=
1
log
dy
y
x
x
dx
⋅
+
⋅
So
dv
dx =
log
y
dy
v
x
x
dx
+
=
log
y
y
dy
x
x
x
dx
+
(3)
Again
w = xx
Taking logarithm on both sides, we have
log w = x log x Differentiating both sides w |
1 | 2316-2319 | x, we have
1 dv
v dx
⋅
=
(log )
log
d
dy
y
x
x
dx
dx
+
=
1
log
dy
y
x
x
dx
⋅
+
⋅
So
dv
dx =
log
y
dy
v
x
x
dx
+
=
log
y
y
dy
x
x
x
dx
+
(3)
Again
w = xx
Taking logarithm on both sides, we have
log w = x log x Differentiating both sides w r |
1 | 2317-2320 | (3)
Again
w = xx
Taking logarithm on both sides, we have
log w = x log x Differentiating both sides w r t |
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