Question

√sinx+cosx differentiate with respect to x
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Answer 1

$y = \sqrt{sinx \:} + cosx \\ \frac{dy}{dx } = \frac{d(\sqrt{sinx \:} + cosx )}{dx} \\ \frac{dy}{dx} = \frac{1}{2} sinx ^{ - \frac{1}{2} } \frac{d(sinx)}{dx} - sinx \\ \frac{dy}{dx} = \frac{1}{2} sinx ^{ - \frac{1}{2} } cosx - sinx$

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