Question

If \( x = \sin t \) and \( y = \tan t \), then \( \frac{dy}{dx} = \)

• A. \( \cos^3 t \)
• B. \( \frac{1}{\cos^3 t} \)
• C. \( \frac{1}{\cos^2 t} \)
• D. \( \sin^2 t \)
• E. \( \frac{1}{\sin^2 t} \)

Hint:

You can also write this as \( \sec^3 t \). When dealing with trig parameters, it's often helpful to convert everything to sine and cosine at the very end to match the given options.

Answer 1

The Correct Option is B

Concept: When \( x \) and \( y \) are given as functions of a parameter \( t \), we use Parametric Differentiation: \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Step 1: Differentiate \( x \) and \( y \) with respect to \( t \).
Given \( x = \sin t \): \[ \frac{dx}{dt} = \cos t \] Given \( y = \tan t \): \[ \frac{dy}{dt} = \sec^2 t = \frac{1}{\cos^2 t} \]

Step 2: Combine the derivatives.
\[ \frac{dy}{dx} = \frac{\frac{1}{\cos^2 t}}{\cos t} \]

Step 3: Simplify the expression.
\[ \frac{dy}{dx} = \frac{1}{\cos^2 t \cdot \cos t} \] \[ \frac{dy}{dx} = \frac{1}{\cos^3 t} \]