A function is defined impilcitly by the equations. 7y2+6x3 Cos(x4y²) - 4 ln(y+5)= 65+ (x³ +7)³ a. Find an expression for at the point (x,y). dx b. Using Part (a) evaluate the derivative at the Point (-2,-4).
A function is defined impilcitly by the equations. 7y2+6x3 Cos(x4y²) - 4 ln(y+5)= 65+ (x³ +7)³ a. Find an expression for at the point (x,y). dx b. Using Part (a) evaluate the derivative at the Point (-2,-4).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Question
A function is defined impilcitly by the equations.
7y2+6x3 Cos(x - y²) - 4 ln(y + 5) = 65+ (x³ +7)³
a. Find an expression for
at the point (x,y).
dx
b. Using Part (a) evaluate the derivative at the Point (-2,-4).
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Follow-up Questions
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Follow-up Question
Hi,
I believe you may have forgotten to multiply 3x^2cos(x^4-y^2) by the 6?
Giving 18x^2cos(c^4-y^2)
See highlighted in image
![Step 2: Calculation:
(a) Given the equations:
7y² + 6x³ cos(x² - y²) - 4 ln (y+5)=65+ (x²¾+7) 3
Differentiating both sides of w.n.tex' we get
dx [74 276 x ³ cos(x 4-y²)-4(y+5)] = [65+ (x²=7)³]
>7.0 (y²)+6.x (x² cos(x 2 y 3) -4.2x (In(y+5))
dx
= 0x (65) + ((x²+7)³)
dx
dx
7.24d/ +6 [x³ (cos(x²¼ y³) + cos(x²±43). ox (x3)]
dx
-
dx
dx
· 4. 115 3 x ( 1 + 5) = 0 + 3(x² + 7)² (x²¾7)
y+5
dy
dx
⇒ 14y diy / +6 [x³ (-sin(x² y²). (4x²-24 1/2) + cos(x²-4³)-3x²)
dy
dx
- 4 (3/2+0) = 3(x+7)²= (3x²+0)
>14 y d²/ +6 [ -x³sin (x²≤ y²) (4x³ 2 ydy 1) + 3x cos(x²yz)]
dx
2
- 4 dy = 9x² (x²³+7)²
y+52x
→ 144 dy - 24 x 6 sin(x²-43) +122³y dy sin(x²-42)
dy-24x6sinlay)+1225y
+3x² cos(x² - y²) -
4 dy = 9x² (x²+7)²
4+5x
dy [144 +12x³ y sin (x²1 y2) - 45]
dy
=
→ कट
--
y+5]
24x6 Sin (x²-42) -3x² cos(x²¼ y²) +9x² (x²³+7)²
24x6 sin(x-2)-3x² cos(x²+ y²)+9x² (x³+7)²
144 +12x³ y sin (x 4_42) - 4
y+5
(b)[22] - 24(-2) 6 sin (0) -3(-2)²-c05(0)+3(-2)² (-8 +7) 2
=
Tat(-254)
1
=
Կ
14(-4)+12 (2)³ (-4) Sin (0) - 4/4+5
0-12 (1) +9 (4)(-1)²
-56+0-
-12+36(1)
-12+36
24
-56-4
- во
.60
=
#P
Solution
(a) dy - 24x6 sin(x4y²) -3x²-COS (x²-4²) +9x² (x³+7)2
dx
=
144 +12x³y sin(x4_43) -
45
4+5](https://content.bartleby.com/qna-images/question/883f8338-cc9c-4943-8997-108926cc85f9/b5310fb9-bc20-4917-baf3-8129088d1897/xhpfjoa_processed.png)
Transcribed Image Text:Step 2: Calculation:
(a) Given the equations:
7y² + 6x³ cos(x² - y²) - 4 ln (y+5)=65+ (x²¾+7) 3
Differentiating both sides of w.n.tex' we get
dx [74 276 x ³ cos(x 4-y²)-4(y+5)] = [65+ (x²=7)³]
>7.0 (y²)+6.x (x² cos(x 2 y 3) -4.2x (In(y+5))
dx
= 0x (65) + ((x²+7)³)
dx
dx
7.24d/ +6 [x³ (cos(x²¼ y³) + cos(x²±43). ox (x3)]
dx
-
dx
dx
· 4. 115 3 x ( 1 + 5) = 0 + 3(x² + 7)² (x²¾7)
y+5
dy
dx
⇒ 14y diy / +6 [x³ (-sin(x² y²). (4x²-24 1/2) + cos(x²-4³)-3x²)
dy
dx
- 4 (3/2+0) = 3(x+7)²= (3x²+0)
>14 y d²/ +6 [ -x³sin (x²≤ y²) (4x³ 2 ydy 1) + 3x cos(x²yz)]
dx
2
- 4 dy = 9x² (x²³+7)²
y+52x
→ 144 dy - 24 x 6 sin(x²-43) +122³y dy sin(x²-42)
dy-24x6sinlay)+1225y
+3x² cos(x² - y²) -
4 dy = 9x² (x²+7)²
4+5x
dy [144 +12x³ y sin (x²1 y2) - 45]
dy
=
→ कट
--
y+5]
24x6 Sin (x²-42) -3x² cos(x²¼ y²) +9x² (x²³+7)²
24x6 sin(x-2)-3x² cos(x²+ y²)+9x² (x³+7)²
144 +12x³ y sin (x 4_42) - 4
y+5
(b)[22] - 24(-2) 6 sin (0) -3(-2)²-c05(0)+3(-2)² (-8 +7) 2
=
Tat(-254)
1
=
Կ
14(-4)+12 (2)³ (-4) Sin (0) - 4/4+5
0-12 (1) +9 (4)(-1)²
-56+0-
-12+36(1)
-12+36
24
-56-4
- во
.60
=
#P
Solution
(a) dy - 24x6 sin(x4y²) -3x²-COS (x²-4²) +9x² (x³+7)2
dx
=
144 +12x³y sin(x4_43) -
45
4+5
Solution
by Bartleby ExpertFollow-up Question
From line 2 to 3 (in step 2 calculations) where did the 6 go for 6 d/dx [x^3cos(x^4-y^2)]
![7[2] + [x³-
3
dx
d
7y²+6 [r³cos(x²— y²)] − 4——|_ ln(y+5) =
7|2y
dy
dx
dx
d
-
d
d
d
4.
65+
dx
dx
+7)³
3
dx
-cos(x4 - y²) + cos(x4 — y²).
d
1
dy
-x· 4.
dx
(y+5) dx
·=0+ 3(x³ +7) ²(3x²)](https://content.bartleby.com/qna-images/question/883f8338-cc9c-4943-8997-108926cc85f9/843fba24-2291-4af5-8d74-f668df0e293e/0w3dbk_processed.png)
Transcribed Image Text:7[2] + [x³-
3
dx
d
7y²+6 [r³cos(x²— y²)] − 4——|_ ln(y+5) =
7|2y
dy
dx
dx
d
-
d
d
d
4.
65+
dx
dx
+7)³
3
dx
-cos(x4 - y²) + cos(x4 — y²).
d
1
dy
-x· 4.
dx
(y+5) dx
·=0+ 3(x³ +7) ²(3x²)
Solution
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