Question Detail

MTH 102 MTH 102 Past Questions Objective Question

Question

Differentiate the implicit function x2 siny - e2y cosx = 3x2 + 2x - 5 with respect to x.
Options
A <i>dy/dx = (6x - 2x siny + e<sup>2y</sup> sin x + 2) / (x<sup>2</sup> cosy + 2e<sup>2y</sup> cosx)</i>
Correct Answer
B <i>dy/dx = (6x + 2x siny - e<sup>2y</sup> sin x + 2) / (x<sup>2</sup> cosy - 2e<sup>2y</sup> cosx)</i>
C <i>dy/dx = (6x - 2x siny - e<sup>2y</sup> sin x + 2) / (x<sup>2</sup> cosy - 2e<sup>2y</sup> cosx)</i>
D <i>dy/dx = (-6x - 2x siny - e<sup>2y</sup> sin x - 2) / (x<sup>2</sup> cosy - 2e<sup>2y</sup> cosx)</i>
Correct Answer

Option A is the correct answer.

Detailed Explanation

Differentiate each term with respect to x using the product rule and chain rule:1. d/dx(x2 siny) = 2x siny + x2 cos y (dy/dx)2. d/dx(-e2y cosx) = -[ (d/dx(e2y)) cosx + e2y (d/dx(cosx)) ] = -[ (2e2y (dy/dx)) cosx + e2y (-sin x) ] = -2e2y cosx (dy/dx) + e2y sin x3. d/dx(3x2 + 2x - 5) = 6x + 2Combine these into the original equation:2x siny + x2 cos y (dy/dx) + 2e2y cosx (dy/dx) - e2y sin x = 6x + 2Group terms with dy/dx on one side and other terms on the other:(x2 cos y + 2e2y cosx) (dy/dx) = 6x + 2 - 2x siny + e2y sin xdy/dx = (6x + 2 - 2x siny + e2y sin x) / (x2 cos y + 2e2y cosx)This matches option A (note that e2y sin x can be written as sin x e2y).

Hint

Use implicit differentiation. Remember the product rule for terms like x2siny and e2ycosx, and the chain rule for derivatives involving y.

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