Higher_Mathematics_Part_1
.pdfMicromodule 18
BASIC THEORETICAL INFORMATION
DERIVATIVE AND ITS CALCULATION (CONTINUED)
Derivative of inverse function. Derivative of implicit function and function in parametric form. Logarithmic differentiation.
Literature: [3, chapter 3, §§ 3.1—3.8], [4, part 5], [6, chapter 5, § 2], [9], [10, chapter 4, § 5], [11, chapter 4, § 2], [12, chapter 3, §§ 16—18].
18.1. Derivative of inverse function
Let y = f (x) and x = g( y) be a pair of inter-inverse functions (remember,
that the plots of such functions coinside).
Let’s prove a theorem about connection between derivatives of these
functons. |
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If functon y = f (x) |
is monotone on the interval |
(a; b) and |
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Theorem 3.10 |
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has non-zero derivative |
f ′(x) at any point of this interval, then |
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there exists inverse function x = g( y), which has derivative x′ = g( y), and |
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g′( y) = |
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or y′ = |
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f |
′(x) |
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x′ |
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This formula has geometrical interpritation. |
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The curve is set by the function y = f (x) |
or inverse function x = g( y) |
(Fig. 3.15). |
Then f ′(x) = tg α ( α is the angle between the tangent line and x-axis),
g′( y) = tg β
Then, tg α =
( β is the angle between the tangent line and y-axis). Since α + β = |
π . |
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π |
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tg ( |
− β) = ctgβ = |
and y′ = |
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2 |
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tgβ |
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x′ |
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y
y= f(x) x = g(x)
M(x; y)
α
О β |
x |
Fig. 3.15
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18.2. Differentiation of implicit function |
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Let the implicit function |
y = f (x) is set by equation |
F(x, y) = 0. In order |
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to find the |
derivative y′(x) , |
we need to differentiate both parts of equation |
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F(x, y) = 0 |
with respect to |
x , and not to forget that |
y is |
a function of |
variable x . The obtained equation may be solved with respect to |
′ |
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y (x) . So, we |
find derivative from the condition
dxd F(x, y) = 0.
Derivative of implicit function is expressed through the independent variable x and function y itself.
18.3. Differentiation of functions given parametrically
The derivative of function y = f (x), given by parametrical equation x = x(t),
y = y(t) (where x(t) and y(t) |
are differentiable at the point t and y′(t) = 0, is |
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found by the formula |
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d |
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dy |
ψ′(t) or y′ |
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yt′ |
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= |
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dt |
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dx |
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dx |
ϕ′(t) |
x |
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dt |
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18.4. Logarithmic differentiation
In some cases while finding derivative we should at first take the logarithm of given function and then find the derivative as in the case of implicit functon. This operation is called logarithmic differentiation.
It is particularly convenient to use logarithmic differentiation if the function is represented as:
а) y = u1k1 (x) u2k2 (x) … umkm (x) (x) ; b) or y = u(x)v(x) . v1l1 (x) v2l2 (x) … vmln (x) (x)
Let’s show how to find the derivative of the function y = u(x)v(x) , where
u(x), v(x) are differentiable functions with respect to |
x, u(x) > 0 . |
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Applying the logarithm, we receive |
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ln y = ln(uv ) = v ln u ; |
(ln y)′ = (v ln y)′ ; |
y′ = v′ lnu + v |
u′ ; |
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y |
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u |
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y′ = uv v′ ln u + v |
1 |
u′ = uv ln u |
v′ + vuv−1 u′ |
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Micromodule 18
EXAMPLES OF PROBLEMS SOLUTION
Example 1. Find the derivative |
y′ , if |
x = y3 + 3y . |
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Solution. We receive |
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x′ = 3y2 + 3 ; y′ = |
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x |
x′ |
3( y |
+ 1) |
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y |
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Example 2. Find the derivative |
x′ |
, if |
y = x + ex . |
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y |
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Solution. The functon y = x + ex |
is monotone for |
x R. y′ = 1+ ex > 0 . |
That is why for function y(x) there exists inverse functon x = x( y) and its
derivative is |
x′ = |
1 |
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1+ ex |
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y′ |
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Example 3. Prove that |
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(arcsin x)′ = |
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1− x2 |
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Solution. |
The function y = arcsin x |
(where x [−1; 1] ), is inverse to |
function x = sin y , y −π |
2 |
, π |
2 |
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x = sin y increases and the derivative theorem are fullfiled. We receive
(arcsin x)′ = |
1 |
= |
cos y |
On interval |
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−π |
2 |
, π |
2 |
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the function |
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x′ = cos y > 0. That is, all conditions of
1 |
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1 − sin2 y |
1 − x2 |
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Example 4. Find the derivative y′, if x2 + y2 = 1.
Solution. We have 2x + 2yy′ = 0 , and we get y′ = − xy .
Example 5. Find the derivative y′, if x2 + 2xy − y2 = 2x.
Solution. Let’s differentiate both sides of equation with respect to х keeping in mind that у is a function of х:
2x + 2( y + xy′) − 2yy′ = 2 , or x + y + xy′ − yy′ = 1 .
We find y′(x − y) = 1− x − y. That is, |
y′ = |
1 − x − y . |
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x − y |
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Example 6. Find |
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y′ , if |
arctg |
y |
= ln |
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x2 + y2 . |
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Solution. We have |
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y ′ |
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arctg |
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= |
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ln(x |
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, |
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= |
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(x |
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2 |
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y 2 |
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(x2 + y2 ) |
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dx |
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1+ |
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x2 |
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y′ − y |
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2x + 2yy′ |
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= |
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, |
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y |
′x − y = x + yy′, y′(x − y) = x + y, |
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x2 + y2 |
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x2 |
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so |
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x + y |
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y′ = |
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x − y |
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Example 7. Find |
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′ |
, if |
y = a sin t . |
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x |
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= b cos t |
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Solution. y′(t) = a cost , |
x′(t) = −b sin t , then |
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y′ |
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a cos t |
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= − |
a |
ctg(t) . |
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− b sin t |
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Example 8. Find |
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y |
′ |
, if |
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= t |
2 |
− 2t |
3 |
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= 2t + t |
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Solution. Let’s find the derivatives |
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y′(t) = 2t − 6t 2 , x′(t) = 2 + 2t , |
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then |
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y′ |
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2t |
− 6t |
2 |
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− 3t |
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2 + 2t |
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y |
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Example 9. Find |
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y |
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x |
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x = ln (1+ t2 ) |
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Soluton. We have |
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+ t 2 |
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214 |
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http://vk.com/studentu_tk, http://studentu.tk/
Example 10. Find the derivative of function |
y = |
x2 |
(3x − 2)ex |
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sin x |
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2x + 1 |
Solution. We use logarithmic differentiation and receive:
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x2 (3x − 2)ex |
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ln y = 2 ln x + ln(3x − 2)+ x − ln sin x − |
1 |
ln(2x + 1); |
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ln y = ln |
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sin x 2x + 1 |
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ln y = 2 ln x + ln(3x − 2)+ x − ln sin x − 1 ln(2x + 1); |
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cos x − |
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x |
3x − 2 |
sin x |
2x + 1 |
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or |
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y′ = |
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− ctg x − |
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sin x 2x + 1 |
3x − 2 |
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Example 11. Find the derivative of function y = xcos x .
Solution. First method. Let’s use logarithmic differentiation. We have
ln y = ln xcos x ; |
ln y = cos x ln x; |
(ln y)′ = (cos x ln x)′; |
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y′ |
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1 |
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= − sin x ln x |
+ cos x |
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y′ = y |
− sin x ln x + cos x |
x |
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That is,
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y′ = x |
cos x |
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Second method. Using basic logarithmic equality |
aloga b = b, |
let's write |
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the given function as |
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y = xcos x |
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= ecos xln x . |
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Then |
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y′ = (ecos xln x )′ = ecos xln x (cos x ln x)′ = |
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= e |
cos xln x |
− sin x ln x + cos x |
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cos x |
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− sin x ln x + cos x |
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215 |
http://vk.com/studentu_tk, http://studentu.tk/
Micromodule 18
CLASS AND HOME ASSIGNMENT
Find dydx using the rule of differentiation of inverse function, if:
1. |
x = y2 + |
y2 + 1 . |
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2. |
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3. |
x = lg cos y + cos ln y. |
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4. |
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x = earccos y . |
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Find the derivative |
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of implicit functions: |
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5. 3x+y |
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6. |
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arctg |
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x2 + y2 . |
7. x y = yx . |
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8. x |
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9. |
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Find the derivative |
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of functions given parametrically: |
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10. |
x = a cos2 t , y = b sin2 t . |
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11. x = t , |
y = 3 t . |
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12. |
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13. |
x = et , |
y = e2t . |
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14. x = et |
cos t , |
y = et |
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15. |
x = a(t −sin t) , |
y = a(1−cos t) . |
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x = t2 , |
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x = e2t |
cos2 t , |
y = e2t |
sin2 t . |
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18. |
x = t cos t , |
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19. |
x = |
cos3 t |
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sin3 |
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Find the derivative |
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20. |
y = (ln x)x . |
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21. |
y = (2x +1)2x−1 . |
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22. y = |
2x tg x 5 x |
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23. |
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y = (x5 +5 |
x )arctg x . |
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25. |
y = xln x +(ln x)x . |
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y = (sin x)cos x (cos x)sin xx . |
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27. |
y = (x5 +5 |
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28. |
y = xxx . |
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216
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Answers
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1. |
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y2 + 1 |
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3x (1− 3y ) |
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x x2 + y2 + y |
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y(2 |
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ln y +1+ cos y |
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3y (1+ 3x ) |
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8. |
y − x2 |
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−b / a . 11. |
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12. t(2 − t3 ) . 13 2et . 14. |
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16. |
2sin2 t + sin 2t . |
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ln y − y / x ln x − x / y .
sin t + cost . cost − sin t
Micromodule 18
SELF-TEST ASSIGNMENT
18.1. Find the derivative |
dy |
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18.1.1. x2 y + y2 x = x3 y3 . |
18.1.2. y = arctg x −arctg y . |
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18.1.3. sin(xy) = x2 + y2 . |
18.1.4. y cos x = sin(x −y) . |
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18.1.5. 3x +3y = 3x+y . |
18.1.6. x3 + y3 −4axy = 0 . |
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18.1.7. ln(x + y) + x2 y =1. |
18.1.8. x sin y = x2 + y2 . |
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18.1.9. x = y3 −4 y +1. |
18.1.10. sin x −cos y = x −y . |
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18.1.11. cos(xy) +sin(xy) = y . |
18.1.12. ctg 2y = 2 ctg x . |
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18.1.13. y3 + x3 y + xy2 =1 . |
18.1.14. x4 + y4 = x3 y3 . |
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18.1.15. y = x −arcsin y . |
18.1.16. x3 y + y3 x = x −y2 . |
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18.1.17. x2 y2 +2xy + x3 = y3 . |
18.1.18. sin(x + y) = x −y . |
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18.1.19. x3 + y4 x = x3 −2y . |
18.1.20. x tg y − y tg x = yx . |
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18.1.21. x3 y −y3 x = (x −y)3 . |
18.1.22. arctg(x + y) = x −y2 . |
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18.1.23. 5x −5y = 5x+y . |
18.1.24. y sin x + x sin y = y . |
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18.1.25. 3y ln y = x2 ( y +5) . |
18.1.26. y3 −5y +6ax = 0 . |
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18.1.27. x3 y2 +2x−y = y . |
18.1.28. 3x+y +3x−y = y3 . |
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18.1.29. y = x +e1+xy . |
18.1.30. arcsin(x / y) + yx = y . |
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18.2. Find the derivative |
dy |
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18.2.2. y =1/ cos2 t , x = ln tg t. |
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y = arccost, |
x = arcsin t. |
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18.2.3. |
y = bsin3 t, |
x = a cos3 t. |
18.2.4. y =1/ sin2 t , x = ln ctg t. |
217
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18.2.6. y = e |
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18.2.8. y = et |
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x = et |
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18.2.9. y = |
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18.2.10. y = |
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18.2.11. y = a(1−cos 2t) , |
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18.2.12. y = a(sin t −t cos t) , |
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18.2.13. y = arcsin(t −1) , x = |
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18.2.14. y = t |
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18.2.15. y = 3 ln tg t , x = |
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18.2.16. y = t −arctgt , x = ln ctgt . |
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18.2.17. y = |
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18.2.18. y = |
tg t |
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x = |
ctg 2t |
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sin2 t |
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18.2.19. y = ln |
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18.2.20. y = |
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18.2.24. y = 1+ t2 −1 , |
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18.2.26. y = t −arctg |
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18.2.28. y = arctg |
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18.2.29. y = tg3 2t + ctg3 2t , |
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18.2.30. y = arcsin |
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18.3. Find the derivative dydx rentiation.
18.3.1. y = xarcsin x . 18.3.3. y = (x3 +1)sin x .
18.3.5. y = (sin x )1/ x . 18.3.7. y = (ctg 5x)5x−1 . 18.3.9. y = xe−tg x .
18.3.11. y = (x − 5)2 3 x2 + 1 . ex (x + 2)5
18.3.13. y = (x − 1)3 4 x2 + x .
4x (3x − 2)3
18.3.15. y = (x3 −x)x2 +1 . 18.3.17. y = xarctg x .
18.3.19. y = (x cos x)ln x . 18.3.21. y = (tg x)ctg x .
18.3.23. y = (cos x)tg x . 18.3.25. y = (4x −3)arccos x . 18.3.27. y = (ctg 2x)ctg x . 18.3.29. y = (5x +2)sin x .
of functions using the rule of logarithmic diffe-
18.3.2. |
y = (lg x)x / 2 . |
18.3.4. |
y = (cos 2x)ln tg x / 2 . |
18.3.6. |
y = xex . |
18.3.8. |
y = (x5 +1)ctg x . |
18.3.10. |
y = (x8 + 1)tg x . |
18.3.12. |
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18.3.16. y = (2x −3)cos x .
18.3.18. y = (x sin x)x2 .
18.3.20. y = x2x . 18.3.22. y = (arcsin x)sin x .
18.3.24.y = x4x .
18.3.26.y = (ln(x +1))ln2 x .
18.3.28.y = xsin x +(sin x)x .
18.3.30. y = xarctg x .
Micromodule 19
BASIC THEORETICAL INFORMATION
DIFFERENTIAL OF FUNCTION. TANGENT
Differential of function. Geometrical interpretation of differential. The application of differentials in calculus of approximation. Tangent and normal.
Literature: [3, chapter 3, §§ 3.1—3.8], [4, section 4], [6, chapter 5, § 3], [7, chapter 6, § 17], [9], [10, chapter 4, § 14], [11, chapter 4, § 3], [12, chapter 3, §§ 20, 26].
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19.1. Definition and geometrical interpretation of differential
Let function y = f (x) be differentiable at the point x, so that it has the derivative at this point:
f ' (x) = lim |
y |
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In general case f ' (x) ≠ 0 . Then |
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yx = f ' (x) + α , where α → 0 if x → 0 , therefore the increment of the function is
y = f '(x) x + α x.
The first term is linear with respect to x , the second one is infinitesimal of the higher order than x . Therefore the first term forms the main part of the increment of function, which is called the differential of function.
Definition 3.21. The main, linear with respect to |
x , part of the increment |
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f (x) is called the differential dy of function y = f (x) : |
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dy = f ′(x) |
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Differential dy |
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If y = x |
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x , so the differential of the independent |
variable x is equal to its increment. Therefore:
dy = f ′(x)dx .
The geometrical interpretation of the differential is clear from Fig.3.16.
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Fig. 3.16
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