Question

y = Sinx. Cosex then dy / dx

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Answer 1

Given :-

• $\large \tt \: y = Sinx.Cosx$

To Find :-

• $\large \tt\frac{dy}{dx}$

Solution :-

$\implies \tt \: y = Sinx.Cosx$

$\tt \blue{ {Differentiate \: with \: respect \: to \: \red{x}.}}$

$\implies \tt \frac{dy}{dx} = \frac{d}{dx} (sinx.cosx)$

$\tt \: \green{Applying \: chain \: rule:-}$

$\tt \: if \: y = pq \: , \: then$

$\implies \tt \: \frac{dy}{dx} = p( \frac{dq}{dx}) + q( \frac{dp}{dx} )$

$\implies \tt \: \frac{dy}{dx} = \frac{d}{dx} (sinx.cosx)$

$\implies \tt \: \frac{dy}{dx} = sinx[\frac{d}{dx} (cosx)] + cosx[ \frac{d}{dx} (sinx)]$

$\implies \tt \: \frac{dy}{dx} = sinx \times ( - sinx) + cosx(cosx)$

$\implies \tt \: \frac{dy}{dx} = cos {}^{2} x - sin {}^{2} x$

$\tt \: We \: know \: that, \orange{[cos2x = cos {}^{2} x - sin {}^{2} x]}$

Hence,

$\huge \pink{ \boxed{\tt \: \frac{dy}{dx} = cos2x}}$