Question

Find dy/dx of the following y = sinx.tanx​

Answer 1

Explanation:

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Answer 2

$y = sinx.tanx \\ \\ \frac{dy}{dx} = sinx \frac{d}{dx} tanx + tanx \frac{d}{dx} sinx \\ \\ \frac{dy}{dx} = sinx {sec}^{2} x + tanxcosx \\ \\ \frac{dy}{dx} = tanxsecx + sinx$