Let g(x)=(cos(x))/(sin(x)). Find g^(')(x). Choose 1 answer: (A) (sin^(2)(x)-cos^(2)(x))/(sin^(2)(x)) (B) (1)/(sin^(2)(x)) (C) -(1)/(sin^(2)(x)) (D) (cos^(2)(x)-sin^(2)(x))/(sin^(2)(x))

Recognize Quotient Rule: Recognize that g(x) = (cos(x))/(sin(x)) is a quotient of two functions, so we will use the quotient rule to find its derivative. The quotient rule states that if h(x) = f(x)/g(x), then h'(x) = (f'(x)g(x) - f(x)g'(x))/(g(x))^2.Identify Functions: Identify the functions f(x) and g(x) where f(x) = cos(x) and g(x) = sin(x). We will need to find the derivatives f'(x) and g'(x).Find f'(x): Find the derivative of f(x) = cos(x). The derivative of cos(x) with respect to x is -sin(x). So, f'(x) = -sin(x).Find g'(x): Find the derivative of g(x) = sin(x). The derivative of sin(x) with respect to x is cos(x). So, g'(x) = cos(x).Apply Quotient Rule: Apply the quotient rule. Using the derivatives from steps 3 and 4, we get: g'(x) = (-sin(x) * sin(x) - cos(x) * cos(x)) / (sin(x))^2.Simplify Expression: Simplify the expression. We can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to simplify the numerator: g'(x) = (-sin^2(x) - cos^2(x)) / (sin(x))^2 = -(sin^2(x) + cos^2(x)) / (sin(x))^2 = -(1) / (sin(x))^2.Recognize Final Answer: Recognize that -(1) / (sin(x))^2 is the same as -(1) / (sin^2(x)), which matches answer choice (C).

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Let
g(x)=(cos(x))/(sin(x)).
Find
g^(')(x).
Choose 1 answer:
(A)
(sin^(2)(x)-cos^(2)(x))/(sin^(2)(x))
(B)
(1)/(sin^(2)(x))
(C)
-(1)/(sin^(2)(x))
(D)
(cos^(2)(x)-sin^(2)(x))/(sin^(2)(x))

Let $g(x)=\frac{\cos (x)}{\sin (x)}$ . $\newline$ Find $g^{\prime}(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{\sin ^{2}(x)-\cos ^{2}(x)}{\sin ^{2}(x)}$ $\newline$ (B) $\frac{1}{\sin ^{2}(x)}$ $\newline$ (C) $-\frac{1}{\sin ^{2}(x)}$ $\newline$ (D) $\frac{\cos ^{2}(x)-\sin ^{2}(x)}{\sin ^{2}(x)}$

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Math Problems

Calculus

Find derivatives of using multiple formulae

Full solution

Q. Let

g(x)=\frac{\cos (x)}{\sin (x)}

\newline

Find

g^{\prime}(x)

\newline

Choose

1

answer:

\newline

(A)

\frac{\sin ^{2}(x)-\cos ^{2}(x)}{\sin ^{2}(x)}

\newline

(B)

\frac{1}{\sin ^{2}(x)}

\newline

(C)

-\frac{1}{\sin ^{2}(x)}

\newline

(D)

\frac{\cos ^{2}(x)-\sin ^{2}(x)}{\sin ^{2}(x)}

Recognize Quotient Rule: Recognize that $g(x) = \frac{\cos(x)}{\sin(x)}$ is a quotient of two functions, so we will use the quotient rule to find its derivative. The quotient rule states that if $h(x) = \frac{f(x)}{g(x)}$ , then $h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$ .
Identify Functions: Identify the functions $f(x)$ and $g(x)$ where $f(x) = \cos(x)$ and $g(x) = \sin(x)$ . We will need to find the derivatives $f'(x)$ and $g'(x)$ .
Find $f'(x)$ : Find the derivative of $f(x) = \cos(x)$ . The derivative of $\cos(x)$ with respect to $x$ is $-\sin(x)$ . So, $f'(x) = -\sin(x)$ .
Find $g'(x)$ : Find the derivative of $g(x) = \sin(x)$ . The derivative of $\sin(x)$ with respect to $x$ is $\cos(x)$ . So, $g'(x) = \cos(x)$ .
Apply Quotient Rule: Apply the quotient rule. Using the derivatives from steps $3$ and $4$ , we get: $g'(x) = \frac{-\sin(x) \cdot \sin(x) - \cos(x) \cdot \cos(x)}{(\sin(x))^2}$ .
Simplify Expression: Simplify the expression. We can use the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$ to simplify the numerator: $\newline$ $g'(x) = (-\sin^2(x) - \cos^2(x)) / (\sin(x))^2 = -(\sin^2(x) + \cos^2(x)) / (\sin(x))^2 = -(1) / (\sin(x))^2.$
Recognize Final Answer: Recognize that $-(1) / (\sin(x))^2$ is the same as $-(1) / (\sin^2(x))$ , which matches answer choice $(C)$ .