Question

It’s differentiation, can someone please show the steps and explain, thank you

Answer 1

Given :

$\\ \red\bigstar\bf{Differentiation \: of \: sinx.cosecx}$

To Find :

$\blue\bigstar\bf{Differentiation \: of \: sinxcosx =?}$

Formula :

$\pink \bigstar \tt \: \frac{d}{dx}sinx = cosx \\ \\ \pink \bigstar \tt \: \frac{d}{dx} cosx = - sinx \\ \\ \pink \bigstar \tt \: \frac{d}{dx} cosecx = - cosecx.cotx \\ \\ \pink \bigstar \tt \: \frac{d}{dx} cotx = - {cosec}^{2}x \\ \\ \pink \bigstar \tt\frac{d}{dx} tanx = {sec}^{2} x$

Solution :

According to the given conditions :

• Consider a function y=f(x) in which y=sinx.cosecx

• As we can see the function is composite so we use Product rule .

$\implies \sf \: y = sinxcosecx \\ \\ \implies \bf \bullet \: \: differentiating \: the \: equation \: with \: to \: x \\ \\ \implies \sf \: \frac{dy}{dx} = \frac{d}{dx} sinx.cosecx \\ \\ \implies \sf \: \frac{dy}{dx} = \frac{d}{dx} sinx.cosecx + \frac{d}{dx} cosecx.sinx \\ \\ \implies \sf \frac{dy}{dx} = cosx.cosecx - cosecx.cotx.sinx \\ \\ \implies \sf \: \frac{dy}{dx} = cosx.cosecx - cosecx. \frac{cosx}{sinx} \times sinx \\ \\ \implies \sf \: \frac{dy}{dx} = cosx.cosecx - cosecx.cotx \\ \\ \implies \boxed { \sf{ \frac{dy}{dx} = 0}}$

The correct answer is 0

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Additional information :

$\\ \red\bigstar\bf{ secx=\dfrac{1}{cosx}}$

$\\ \red\bigstar\bf{tanx=\dfrac{1}{cotx}=\dfrac{sinx}{cosx}}$

$\\ \red\bigstar\bf{cosecx=\dfrac{1}{sinx}}$

$\\ \red\bigstar\bf{cotx=\dfrac{1}{tanx}}$

$\\ \red\bigstar\bf{cotx=\dfrac{cosx}{sinx}}$