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A A A A A A A A A A A A A A A A A A A A A A A

If
y=x sin x, then
(dy)/(dx)=
(A)
sin x+cos x
(B)
sin x+x cos x
(C)
sin x-x cos x
(D)
x(sin x+cos x)
(E)
x(sin x-cos x)
Let
f be the function given by
f(x)=300 x-x^(3). On which of the following intervals is the function
f increasing?
(A)
(-oo,-10] and
[10,oo)
(B)
[-10,10]
(C)
[0,10] only
(D)
[0,10sqrt3] only
(E)
[0,oo)
Unauthorized copying or reuse of
any part of this page is illegal.
GO ON TO THE NEXT PAGE.
-4-
A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A

int sec x tan xdx=
(A)
sec x+C
(B)
tan x+C
(C)
(sec^(2)x)/(2)+C
(D)
(tan^(2)x)/(2)+C
(E)
(sec^(2)xtan^(2)x)/(2)+C

A A A A A A A A A A A A A A A A A A A A A A A $\newline$ $1$ . If $y=x \sin x$ , then $\frac{d y}{d x}=$ $\newline$ (A) $\sin x+\cos x$ $\newline$ (B) $\sin x+x \cos x$ $\newline$ (C) $\sin x-x \cos x$ $\newline$ (D) $x(\sin x+\cos x)$ $\newline$ (E) $x(\sin x-\cos x)$ $\newline$ $2$ . Let $f$ be the function given by $f(x)=300 x-x^{3}$ . On which of the following intervals is the function $f$ increasing? $\newline$ (A) $\frac{d y}{d x}=$ $0$ and $\frac{d y}{d x}=$ $1$ $\newline$ (B) $\frac{d y}{d x}=$ $2$ $\newline$ (C) $\frac{d y}{d x}=$ $3$ only $\newline$ (D) $\frac{d y}{d x}=$ $4$ only $\newline$ (E) $\frac{d y}{d x}=$ $5$ $\newline$ Unauthorized copying or reuse of $\newline$ any part of this page is illegal. $\newline$ GO ON TO THE NEXT PAGE. $\newline$ $-4$ - $\newline$ A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A $\newline$ $3$ . $\frac{d y}{d x}=$ $6$ $\newline$ (A) $\frac{d y}{d x}=$ $7$ $\newline$ (B) $\frac{d y}{d x}=$ $8$ $\newline$ (C) $\frac{d y}{d x}=$ $9$ $\newline$ (D) $\sin x+\cos x$ $0$ $\newline$ (E) $\sin x+\cos x$ $1$

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Math Problems

Grade 7

One-step inequalities: word problems

Full solution

Q. A A A A A A A A A A A A A A A A A A A A A A A

\newline

1

. If

y=x \sin x

, then

\frac{d y}{d x}=

\newline

(A)

\sin x+\cos x

\newline

(B)

\sin x+x \cos x

\newline

(C)

\sin x-x \cos x

\newline

(D)

x(\sin x+\cos x)

\newline

(E)

x(\sin x-\cos x)

\newline

2

. Let

f

be the function given by

f(x)=300 x-x^{3}

. On which of the following intervals is the function

f

increasing?

\newline

(A)

\frac{d y}{d x}=

0

and

\frac{d y}{d x}=

1

\newline

(B)

\frac{d y}{d x}=

2

\newline

(C)

\frac{d y}{d x}=

3

only

\newline

(D)

\frac{d y}{d x}=

4

only

\newline

(E)

\frac{d y}{d x}=

5

\newline

Unauthorized copying or reuse of

\newline

any part of this page is illegal.

\newline

GO ON TO THE NEXT PAGE.

\newline

-4

\newline

A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A

\newline

3

\frac{d y}{d x}=

6

\newline

(A)

\frac{d y}{d x}=

7

\newline

(B)

\frac{d y}{d x}=

8

\newline

(C)

\frac{d y}{d x}=

9

\newline

(D)

\sin x+\cos x

0

\newline

(E)

\sin x+\cos x

1

Find Derivative of y: For the first part, we need to find the derivative of $y$ with respect to $x$ , where $y = x \sin x$ . Using the product rule, $(uv)' = u'v + uv'$ , we get: $\frac{dy}{dx} = (x)' \sin x + x (\sin x)'$ .
Calculate Derivatives: Now, we calculate the derivatives: $(x)' = 1$ and $(\sin x)' = \cos x$ . So, $\frac{dy}{dx} = 1 \cdot \sin x + x \cdot \cos x$ .
Determine Function Increase: Therefore, the derivative of $y$ with respect to $x$ is $\sin x + x \cos x$ . So, the correct answer is (B) $\sin x + x \cos x$ .
Find Increasing Interval: Next, we need to find where the function $f(x) = 300x - x^3$ is increasing. $\newline$ To do this, we find the derivative $f'(x)$ and determine where it is positive.
Find Increasing Interval: Next, we need to find where the function $f(x) = 300x - x^3$ is increasing. $\newline$ To do this, we find the derivative $f'(x)$ and determine where it is positive.The derivative $f'(x)$ is $300 - 3x^2$ . $\newline$ We set $f'(x) > 0$ to find where the function is increasing.
Find Increasing Interval: Next, we need to find where the function $f(x) = 300x - x^3$ is increasing. $\newline$ To do this, we find the derivative $f'(x)$ and determine where it is positive.The derivative $f'(x)$ is $300 - 3x^2$ . $\newline$ We set $f'(x) > 0$ to find where the function is increasing.Solving the inequality $300 - 3x^2 > 0$ , we get $x^2 < 100$ . $\newline$ Taking the square root, we find $-10 < x < 10$ .
Find Increasing Interval: Next, we need to find where the function $f(x) = 300x - x^3$ is increasing. To do this, we find the derivative $f'(x)$ and determine where it is positive. The derivative $f'(x)$ is $300 - 3x^2$ . We set $f'(x) > 0$ to find where the function is increasing. Solving the inequality $300 - 3x^2 > 0$ , we get $x^2 < 100$ . Taking the square root, we find $-10 < x < 10$ . Therefore, the function $f$ is increasing on the interval $(-10, 10)$ . But we made a mistake; we should include the endpoints since the derivative is zero there, not undefined.