# Trig Derivatives Memorization Quiz.

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| By KenArrari
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KenArrari
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Quizzes Created: 1 | Total Attempts: 5,415
Questions: 12 | Attempts: 5,492  Settings  Nothing but absolute mindless memorization of the trig derivatives. Keep taking it until you have them memorized.

• 1.

### Y=sin(x)

• A.

Y'= sin(x)

• B.

Y'= cos(x)

• C.

Y'= tan(x)

• D.

Y'=1/(1+x^2)

B. Y'= cos(x)
Explanation
The derivative of sin(x) is cos(x).

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• 2.

### Y= cos (x)

• A.

Y'= sin(x)

• B.

Y'= -sin(x)

• C.

Y'= cot (x)

• D.

Y'= sec (x)

B. Y'= -sin(x)
Explanation
The derivative of the cosine function, y = cos(x), is equal to the negative sine function, y' = -sin(x). This can be derived using the chain rule of differentiation.

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• 3.

### Y= tan(x)

• A.

Y'= csc(x)

• B.

Y'= cot(x)

• C.

Y'= sec^2(x)

• D.

Y'= -cscxCotx

C. Y'= sec^2(x)
Explanation
The derivative of y = tan(x) is y' = sec^2(x). This is because the derivative of tan(x) is equal to the derivative of sin(x)/cos(x), which can be rewritten as 1/cos^2(x). Since 1/cos^2(x) is equivalent to sec^2(x), the correct answer is y' = sec^2(x).

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• 4.

### Y= sec(x)

• A.

Y'= 1/(1+x^2)

• B.

Y'= -csc^2(x)

• C.

Y'= -cos(x)

• D.

Y'= Sec(x)Tan(x)

D. Y'= Sec(x)Tan(x)
Explanation
The derivative of y = sec(x) is y' = sec(x)tan(x). This can be derived using the quotient rule, where the derivative of sec(x) is sec(x)tan(x) and the derivative of tan(x) is sec^2(x). Therefore, the correct answer is y' = sec(x)tan(x).

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• 5.

### Y= cot(x)

• A.

Y'=-csc^2(x)

• B.

Y'= -cos(x)

• C.

Y'= -sin(x)

• D.

Y'= sec(x)tan(x)

A. Y'=-csc^2(x)
Explanation
The derivative of y=cot(x) is y'=-csc^2(x). This is because the derivative of cot(x) can be found using the quotient rule, where the derivative of the numerator is -csc^2(x) and the derivative of the denominator is 1. Therefore, the derivative of cot(x) is -csc^2(x).

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• 6.

### Y= csc(x)

• A.

Y'= Csc(x)

• B.

Y'=-Csc(x)Cot(x)

• C.

Y'= cot(x)tan(x)

• D.

Y'= Sec(x)Tan(x)

B. Y'=-Csc(x)Cot(x)
Explanation
The correct answer is y'=-Csc(x)Cot(x). The derivative of y = csc(x) can be found using the chain rule. The derivative of csc(x) is -csc(x)cot(x), where -csc(x) represents the derivative of the outer function and cot(x) represents the derivative of the inner function. Therefore, the correct answer is y'=-Csc(x)Cot(x).

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• 7.

### Y= Sin^-1(x)

• A.

Y'=1/√(1-x^2)

• B.

Y'=-1/√(1-x^2)

A. Y'=1/√(1-x^2)
Explanation
The given function is y = sin^(-1)(x), which represents the inverse sine function. To find the derivative of this function, we can use the chain rule. The derivative of sin^(-1)(x) is 1/√(1-x^2). Therefore, the correct answer is y' = 1/√(1-x^2).

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• 8.

### Y= Tan^-1(x)

• A.

Y'=1/√(1-x^2)

• B.

Y'= 1/(1+x^2)

• C.

Y'= -1/(1+x^2)

• D.

Y'=1/√(1-x^2)

B. Y'= 1/(1+x^2)
Explanation
The given function is y = Tan^-1(x), which represents the inverse tangent of x. The derivative of the inverse tangent function is 1/(1+x^2). This can be derived using the chain rule and the derivative of the tangent function. Therefore, the correct answer is y' = 1/(1+x^2).

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• 9.

### Y= Sec^-1(x)

• A.

Y'=1/√(1-x^2)

• B.

Y'= 1/(1+x^2)

• C.

Y'= -1/(x√(1-x^2)

• D.

Y'= 1/(x√(1-x^2)

D. Y'= 1/(x√(1-x^2)
Explanation
The given expression is y = Sec^-1(x), which represents the inverse secant function. To find the derivative of this function, we can use the chain rule. The derivative of Sec^-1(x) is equal to 1 divided by the square root of (1 - x^2). Therefore, the correct answer is y' = 1/(x√(1-x^2)).

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• 10.

### Y= Cos^-1(x)

• A.

Y'= 1/√(1-x^2)

• B.

Y'= 1/(1+x^2)

• C.

Y'= -1/(1+x^2)

• D.

Y'= -1/√(1-x^2)

D. Y'= -1/√(1-x^2)
Explanation
The given correct answer is y' = -1/√(1-x^2). This can be derived using the chain rule of differentiation. The derivative of y = Cos^-1(x) with respect to x can be found by differentiating the inverse cosine function. Applying the chain rule, we have y' = -1/√(1-x^2), which matches the given correct answer.

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• 11.

### Y= cot^-1(x)

• A.

Y'= -1/(1+x^2)

• B.

Y'= 1/(1+x^2)

• C.

Y'= -1/√(1-x^2)

• D.

Y'= 1/√(1-x^2)

A. Y'= -1/(1+x^2)
Explanation
The given correct answer is y' = -1/(1+x^2). This can be derived using the chain rule of differentiation. The derivative of cot^-1(x) can be found by differentiating the inverse cotangent function. The derivative of cot^-1(x) is equal to -1/(1+x^2). Therefore, the correct answer is y' = -1/(1+x^2).

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• 12.

### Y= csc^-1(x)

• A.

Y'= -1/(x√(1-x^2)

• B.

Y'= -1/√(1-x^2)

• C.

Y'= 1/(1+x^2)

• D.

Y'= 1/(x√(1-x^2) Back to top