Higher_Mathematics_Part_1
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d 2 y |
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d n − 1 y |
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d 3 y |
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d x 2 |
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d n y |
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d x |
n − 1 |
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d x 3 |
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x t′ |
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d x n |
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x t′ |
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20.4. Higher order derivatives of implicit function |
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Assume that a function |
y = f (x) |
is given implicitly with equality F(x, y) = 0. |
Executing differentiation with respect to x and solving the obtained equation relatively to y′ it is possible to determine the first derivative.
To find the second derivative it is necessary to differentiate the first derivative with respect to x and to put its values into obtained relation. While continuing differentiation the any order derivatives can be determined one after another. All of them will be expressed through the independent variable x and the function y itself.
20.4. Higher order differentials
Definition 3.24. The differential of the first-order differential of function
f (x) is |
called the second-order differential of |
twice |
differentiable |
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function |
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y = f (x) |
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i.e. |
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In general case |
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d 2 y = d(dy). |
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d n y = d(d n−1 y). |
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The differentials of the function |
y = f (x) |
are calculated by the formulas: |
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d |
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y = |
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d |
3 |
y = f |
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, …, |
d |
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y = f |
(n) |
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n |
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f (x)(dx) |
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(x)(dx) |
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(x)(dx) |
If x is some function of a variable t, then
d 2 y = f ′′(x)(dx)2 + f ′(x)x′′(t)(dt)2
and so on. Thus the higher order differentials don’t have invariance property.
Micromodule 20
EXAMPLES OF PROBLEMS SOLUTION
Example 1. Determine the third-order derivative of the function y = x4 − − 2x2 + 3x − 5.
231
Solution. Let’s find the third-order derivative step by step y′ = 4x3 − 4x + 3 , y′′ = 12x2 − 4 , y′′′ = 24x .
Example 2. Determine the second-order derivative of the function
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y = |
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x2 −a2 − |
a2 |
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ln(x + |
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x2 −a2 ) , |
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a = const . |
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Solution. We receive |
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y |
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− a |
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+ x ( x |
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− a |
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a 2 |
(ln (x + |
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− a |
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x |
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)) = |
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(x |
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( |
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))= |
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− a |
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+ x |
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2 x |
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a 2 |
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2 x |
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x + |
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x 2 − a 2 + |
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− |
2(x + |
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x 2 − a 2 + x 2 |
− 2 |
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a 2 ( x 2 − a 2 |
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+ x) |
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= |
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2 x 2 − a 2 |
(x + x 2 − a 2 |
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x 2 − a 2 |
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= |
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2x2 − a2 |
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a2 |
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x2 −a2 |
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−a |
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2 x2 − a2 |
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2 x2 − a2 |
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x2 −a2 |
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1 |
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x |
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y '' = ( |
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x 2 − a 2 ) |
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(x 2 − a 2 )' = |
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2x = |
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2 x2 − a2 |
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2 x 2 − a 2 |
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x2 − a2 |
Example 3. Determine the n-order derivative of the functions:
а) |
y = ex ; b) y = ax ; c) |
y = sin x ; d) y = |
1 |
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Solution. |
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а) y′ = ex , y′′ = ex , y(n) = ex ; |
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b) y′ = ax ln a , y′′ = ax ln2 a , y(n) = ax lnn a ; |
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c) |
y′ = cos x = sin(x + π ) , |
y′′ = − sin x = sin(x + 2 |
π ) , |
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2 |
y |
′′′ = − cos x = sin(x + 3 π ) , y(n) = sin(x + n π ) ; |
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2 |
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d) y′ = − x−2 , y′′ = 2x−3 , |
y′′′ = −2 3x−4 , y(4) = 2 3 4x−5 , …, |
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y(n) = (−1)n 2 3 4 … nx−(n+1) = (−1)n n!x−(n+1) . |
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232
Example 4. Determine the n-order derivative of the function
y = |
3x +1 |
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x2 +2x −3 |
Solution. Let’s decompose this function into partial fractions:
3x +1 |
= |
3x +1 |
= |
A |
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+ |
B |
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A(x + 3) + B(x − 1) |
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x2 +2x −3 |
(x −1)(x +3) |
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x −1 |
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x +3 |
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(x −1)(x + 3) |
So,
3x +1= A(x +3) + B(x −1)
Suppose x = 1, then 4 = 4A, whence A = 1. If x = – 3, then – 8 = – 4B, B = 2. Thus
y = |
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+ |
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x −1 |
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x +3 |
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Finally according to case d) of the previous example, we obtain |
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y(n) = (−1)n n! |
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+ 2(−1)n n! |
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(x − 1)n+1 |
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(x +3)n+1 |
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Example 5. Determine the fifth-order derivative y(5) of the function |
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y = (x2 − 3x + 4)e2x . |
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Solution. Let’s designate u = e2x , |
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v = x2 − 3x + 4 . |
Applying Leibniz's |
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formula we can obtaine |
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y(5) = ((x2 − 3x + 4)e2x )(5) |
= (e2x )(5) (x2 − 3x + 4) + |
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+ C1 (e2x )(4) |
(2x − 3) + C2 (e2x )(3) |
2 . |
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5 |
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5 |
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Here we consider that (x2 − 3x + 4)(n) |
= 0 if n > 2. Then we get |
y(5) = 32e2x (x2 − 3x + 4) + 5 16e2x (2x − 3) + 10 8e2x 2 =
= e2x [32(x2 − 3x + 4) + 80(2x − 3) + 160] = e2x (32x2 + 64x + 48) .
Example 6. Find |
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d 2 y |
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, if x = 2 cos t , y |
= 3sin t . |
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dx2 |
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Solution. We have |
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dy |
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(3sin t)t′ |
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3 cos t |
= − |
3 |
ctg t ; |
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dx |
(2 cos t)′ |
−2 sin t |
2 |
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233
Determine the n-order derivative of function applying Leibniz's formula:
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19. |
y = (x3 − 2x + 5) sin x , |
n = 10 . |
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20. |
y = (x2 + 4x − 3)2x , |
n = 8 . |
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21. |
y = x ln(x2 − 3x + 2) , |
n = 6 . |
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Find |
d 2 y |
of implicit function: |
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dx2 |
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22. |
x3 + y3 = 3xy . |
23. ex + x = ey + y . |
24. cos(x + y) = x . |
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25. |
y2 |
= 2 px . |
26. |
y = sin(x + y) . |
27. ln(x + y) = y − x . |
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28. |
xy |
= y . |
29. |
tg |
y |
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= y . |
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Find |
d 2 y |
of parametric function: |
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dx2 |
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30. y = 3t − t3 , x = 2t − t2 . |
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31. y = et sin t , |
x = et cos t . |
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32. y = ln t , x = t6 . |
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33. y = t2 / 2 , x = arctg t . |
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34. y = cos 2t , x = sin2 t . |
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35. y = ln(1+ t2 ) , x = arcctg t . |
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Answers |
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1. 12x(x3 − 2)2 (11x3 − 4). |
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2. 2x2 (x4 + 3)(x4 + 1)−3/ 2 . |
5. xx (ln x + 1)2 + xx−1. |
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6. −4cos 2x / sin2 2x . |
11. − 9cos3x . |
12. (−1)n 3− x lnn 3 . |
13. (−1)n−1(n − 1)!/ xn. |
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14. |
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n |
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15. |
1 |
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n |
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16. n! . |
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sin(kx + nπ / 2). |
2[cos(x + nπ / 2) − |
3 cos(3x |
+ nπ/ 2)] . |
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17. |
(−1)n n![1/(x + 1)n+1 − 1/(x +2)n+1] . |
18. (−1)n n![2/(x −1)n+1 − 1/(x − 3)n+1] . |
19. (30x2 + 700)cos x − (x3 − 272x + 5)sin x. 20. |
2x ln6 2[ln2 2(x2 + 4x − 3) +16ln2(x + 2) + 42] . |
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21. |
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24(6x |
− 11) |
+ |
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24(6x − 23) |
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22. |
2xy(3xy − x3 − y3 − 1) |
. 23. |
(e y − e x )(e x+ y − 1) |
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(x − |
1)6 |
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(x − 3)6 |
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( y 2 − x)3 |
(e y + 1) |
3 |
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− cos(x + y) |
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p 2 |
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− cos |
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24. |
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25. − |
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2 |
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. 27. |
− 4(x + y) /(x + y − 1) |
3 |
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sin3 (x + y) |
y3 |
4 sin |
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x + y |
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30. |
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3/(4(1− t)) . 31. |
2e−t /(cost − sin t)3 . 32. |
− 1/(6t12 ) . |
33. 3t 4 + 4t 2 + 1 . |
34. 0. |
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35. |
2(1 + t 2 ) . |
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235 |
20.2.3. x = |
1 |
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y = ctg t + t. |
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sin t |
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20.2.4. x = lg sin t, |
y = lg cos t. |
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20.2.5. x = sin lg t, |
y = tg lg t. |
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20.2.6. x = cos t + t sin t, y = sin t − t cos t. |
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20.2.7. |
x = cos 2t + 2t sin 2t, y = sin 2t − 2t cos 2t. |
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20.2.8. x = ln(1+ t2 ), |
y = t − arctg t. |
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20.2.9. x = arcsin et , |
y = |
1− e2t . |
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20.2.10. |
x = sin et , y = cos et . |
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20.2.11. |
x = ln ctg t, |
y = |
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1 |
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sin 2t |
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20.2.12. |
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20.2.13. |
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20.2.14. |
x = arctg et , |
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20.2.15.x = ln(1+ 4t2 ), y = 2t − arctg 2t.
20.2.16.x = sin3 et , y = cos3 et .
20.2.17. |
x = 3t cos t, |
y = 3t sin t. |
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20.2.18. |
x = arc ctg t, |
y = log3 (t2 + 1). |
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20.2.19. |
x = cos 2t − ln ctg t, y = sin 2t. |
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20.2.20. |
x = tg 2t , y = ln cos3 2t . |
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20.2.21. |
x = arccos 2t, y = |
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20.2.22. |
x = arcsin t, |
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20.2.23. |
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20.2.24. |
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20.2.25. |
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20.2.26. |
x = ln(1+ t4 ), y = arctg(t2 ). |
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20.2.27. |
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20.2.28. |
x = tg et , |
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20.2.29. |
x = arctg t, |
y = log2 (t2 + 1). |
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20.2.30. |
x = ln(1+ t6 ), y = arctg(t3 ). |
20.3. Determine the n-order derivative of function applying Leibniz's formula:
20.3.1.y = e2x−1 (x3 − 3x + 4) , n=10.
20.3.2.y = (x3 − 2x + 1) sin 2x , n=12 .
20.3.3.y = 2x (3x3 − 7x + 5) , n=15.
20.3.4.y = (x3 + 2x2 + 3) ln x , n=8.
20.3.5.y = (x2 + 4x − 5) ln x , n=10.
20.3.6.y = (x2 + 5x − 12) 2x , n=9.
20.3.7.y = (x3 + 5x2 − 2) sin x , n=11.
20.3.8.y = (x3 − 2x2 − 3) cos x , n=9.
20.3.9.y = (x2 − 4x2 − x + 3) cos 2x , n=8.
20.3.10.y = (x3 + 2x + 3) ln x , n=7.
20.3.11.y = (x2 − 5x) ln(x + 1) , n=8.
20.3.12.y = (2x2 − 7x) ln(x − 2) , n=10.
20.3.13.y = (4x2 − 9x) ln(x − 2) , n=6.
20.3.14.y = (2x2 − 3x −11) 3x , n=9.
20.3.15.y = (4x2 − x + 8) 4− x , n=10.
20.3.16.y = (2x3 − 4x2 − 1) 5− x , n=7.
20.3.17. y = (x2 + 3x − 7) 6− x , n=8.
20.3.18.y = e− x−1 (x3 − x2 + 2) , n=10.
20.3.19.y = 2− x−1 (x3 − 6x + 3) , n=7.
20.3.20.y = 3− x (2x2 + x + 3) , n=8.
20.3.21.y = (4x3 − x2 − 1) cos 2x , n=10.
20.3.22.y = (3x2 − 4x + 1) cos(x + 1) , n=12.
20.3.23.y = (5x2 − 3x + 2) sin(x − 2) , n=11.
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20.3.24. |
y = (6x3 − 1) sin(x + 3) , n=15. |
20.3.25. |
y = (2 x3 − 4 x − 1) ln( x − 3) , n=10. |
20.3.26.y = (3x3 + 2x + 3) ln(x + 3) , n=8.
20.3.27.y = (x2 − 2x − 1) ln(2x − 1) , n=11.
20.3.28.y = (x3 + 2x − 3) ln(2x + 1) , n=10.
20.3.29.y = e−2x (x3 − 6) , n=8.
20.3.30.y = 2− x (x3 − 4) , n=10.
Micromodule 21
BASIC THEORETICAL INFORMATION
BASIC THEOREMS OF DIFFERENTIAL CALCULUS
Fermat’s, Rolle’s, Lagrange’s, Cauchy’s theorems. Taylor’s, Maclaurin’s formulas. L’Hospital’s rule.
Literature: [4, part 5], [6, chapter 5, §5], [7, chapter 6, §19], [9], [10, chapter 4, §24], [11, chapter 4, §4], [12, chapter 4, §§1-7] .
21.1. Fermat’s, Rolle’s, Lagrange’s, Cauchy’s theorems
Those functions, which have the derivative on the current interval, are distinguished by the properties , which help to investigate the function behavior on the interval of differentiability.
Theorem 3.11 (Fermat’s theorem). Let the function f (x) be continuous on
the interval (a; b) and reaches it’s minimum and maximum values at the point on this interval. Therefore, if there’s a derivative f ′(c) at the point с, then f ′(c) = 0 .
This theorem has trivial geometric interpretation. If at the point x = c function f (x) reaches minimum or maximum values (Fig. 3.17 and 3.18), then the tangent to the graph of this function at the point (c; f (c)) is parallel to the x-axis.
Theorem 3.12 (Rolle’s theorem). If the function f (x) is continuous on the segment [a; b] , has the derivative at the each point on the interval (a; b) and at the ends of the interval reaches the same values f (a) = f (b) , then there’s at least one point c (a; b) such as f ′(c) = 0 .
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The geometric interpretation of this theorem is understood from the Fig. 3.19–3.21.
y |
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Fig. 3.18 |
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Fig. 3.19 |
Fig. 3.20 |
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Fig. 3.21 |
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(Lagrange’s theorem). If the function |
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Theorem 3.13 |
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the segment [a; b] and differentiable at the interval (a; b) , |
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This formula also known as Lagrange’s formula. It can be transform as: f (x + x) − f (x) = f ′(x + θΔx) x , 0 < θ < 1.
The geometric interpretation of Lagrange’s theorem. If the function f (x) is
satisfied conditions of Lagrange’s theorem, then there’s at least one point on this |
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Fig. 3.22 |
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