Given the equation x' = -4yf (x) + sin(y) + 1. Take the derivative for both sides of the equation with respect to x. We obtain dy a. 3x? = -4f(x) – 4y- dx b. 3x2 = -4 dy df(x) + cos(y); dy dx dx dx c. 3x? = -4 + cos(y) + 1 с. Зх? dx d. 3x? = -4 + cos(y) dx dy е. Зx2 = -49 + sin(y) dx dx
Given the equation x' = -4yf (x) + sin(y) + 1. Take the derivative for both sides of the equation with respect to x. We obtain dy a. 3x? = -4f(x) – 4y- dx b. 3x2 = -4 dy df(x) + cos(y); dy dx dx dx c. 3x? = -4 + cos(y) + 1 с. Зх? dx d. 3x? = -4 + cos(y) dx dy е. Зx2 = -49 + sin(y) dx dx
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![Given the equation x
-4yf(x) + sin(y) + 1. Take the derivative for both sides of the equation with
respect to x. We obtain
dy
= -4
df(x)
а. Зx?
(x) – 4y²
dy
+ cos(y)
dx
b. 3x?
-4dy df(x)
dx dx
+ cos(y)
dx
с. Зx?
-4% + cos(y)으 + 1
dx
dx
d. 3x?
42 + cos(y)의
= -
dx
dx
e. 3x? = -4 + sin(y)
dx
dx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a95d37a-6d2c-4845-99e3-2b94485914c1%2F03725efd-fa32-436f-8b63-383aace84e90%2Ffno6b9_processed.png&w=3840&q=75)
Transcribed Image Text:Given the equation x
-4yf(x) + sin(y) + 1. Take the derivative for both sides of the equation with
respect to x. We obtain
dy
= -4
df(x)
а. Зx?
(x) – 4y²
dy
+ cos(y)
dx
b. 3x?
-4dy df(x)
dx dx
+ cos(y)
dx
с. Зx?
-4% + cos(y)으 + 1
dx
dx
d. 3x?
42 + cos(y)의
= -
dx
dx
e. 3x? = -4 + sin(y)
dx
dx
![Find the derivative of f(x) = (6x² + 6x – 4)arctan x
1
1
a. (6x2 + 6x – 4)arctan »
arctan x 6x2+бх—4
1
b. (6x2 + 6x – 4)arctan × (-
-In(6x² + 6x – 4) + arctan x-
1+x2
(12x + 6))
бх2+6х-4
12x+6
1
1
c. (6x? + 6x – 4)arctan x In(arctan(x)).
+ In(бx? + 6х - 4):
С.
бх2+бх-4
arctan x 1+x²
d. (6x² + 6x – 4)arctan x (In(arctan(x)).
12x
1
+ In(бx2 + 6х — 4);
бх2+6х-4
1+arctan? x 1+x²
12x-4
e. (6x? + 6x – 4)arctan × (In(arctan(x)).
+ In(6x² + 6x – 4):
2x
arctan x 1++x2
бх2+бх-4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a95d37a-6d2c-4845-99e3-2b94485914c1%2F03725efd-fa32-436f-8b63-383aace84e90%2Fadjzrwx_processed.png&w=3840&q=75)
Transcribed Image Text:Find the derivative of f(x) = (6x² + 6x – 4)arctan x
1
1
a. (6x2 + 6x – 4)arctan »
arctan x 6x2+бх—4
1
b. (6x2 + 6x – 4)arctan × (-
-In(6x² + 6x – 4) + arctan x-
1+x2
(12x + 6))
бх2+6х-4
12x+6
1
1
c. (6x? + 6x – 4)arctan x In(arctan(x)).
+ In(бx? + 6х - 4):
С.
бх2+бх-4
arctan x 1+x²
d. (6x² + 6x – 4)arctan x (In(arctan(x)).
12x
1
+ In(бx2 + 6х — 4);
бх2+6х-4
1+arctan? x 1+x²
12x-4
e. (6x? + 6x – 4)arctan × (In(arctan(x)).
+ In(6x² + 6x – 4):
2x
arctan x 1++x2
бх2+бх-4
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