Home / AP Calculus BC : 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions – Exam Style questions with Answer- MCQ

AP Calculus BC : 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions – Exam Style questions with Answer- MCQ

Question

Given that 3x − tan y = 4, what is \(\frac{\mathrm{d} y}{\mathrm{d} x}\)  in terms of y?
A \(\frac{\mathrm{d} y}{\mathrm{d} x}=3\sin ^{2}y\)
B \(\frac{\mathrm{d} y}{\mathrm{d} x}=3\cos ^{2}y\)
C \(\frac{\mathrm{d} y}{\mathrm{d} x}=3\cos y\cot y\)
D \(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3}{1+9y^{2}}\)

Answer/ExplanationAns:B

Question

\(\frac{\mathrm{d} }{\mathrm{d} x}(\cot x)\)=
A −tanx
B \(\sec ^{2}x\)
C tanx
D \(-\csc ^{2}x\)

Answer/ExplanationAns:D

Question

\(\frac{\mathrm{d} }{\mathrm{d} x}(\sin x\csc x)=\)
A 0
B 1
C \(-\cot ^{2}x\)
D 2cotx

Answer/Explanation

Ans:A

Question

Below is an attempt to derive the derivative of cscx using the product rule, where x is in the domain of cscx. In which step, if any, does an error first appear?

Step 1: cscx⋅sinx=1

Step 2: \(\frac{\mathrm{d} }{\mathrm{d} x}(\csc x.\sin x)=0\)

Step 3:\(\frac{\mathrm{d} }{\mathrm{d} x}(\csc x).\sin x+\csc x.\cos x=0\)

Step 4:\( \frac{\mathrm{d} }{\mathrm{d} x}(\csc x)=\frac{-\csc x.\cos x}{\sin x}=-\csc x.\cot x\)

A Step 1

B Step 2

C Step 3

D There is no error.

Answer/Explanation

Ans:D

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