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Mathematics Question on Differentiability

The derivative of sin⁡x∘  cos⁡x\sin x^{\circ} \, \, \cos x with respect to xx is

A

π180(cos⁡ x∘ cos⁡ x−sin⁡ x∘sin⁡ x)\frac{\pi}{180} (\cos \, x^{\circ} \, \cos \, x - \sin \, x^{\circ} \sin \, x)

B

π180cos⁡ x∘ cos⁡ x−sin⁡ x∘sin⁡ x\frac{\pi}{180} \cos \, x^{\circ} \, \cos \, x - \sin \, x^{\circ} \sin \, x

C

180π(cos⁡ x∘ cos⁡ x−sin⁡ x∘sin⁡ x)\frac{180}{\pi} (\cos \, x^{\circ} \, \cos \, x - \sin \, x^{\circ} \sin \, x)

D

−π180cos⁡ x∘ sin⁡ x- \frac{\pi}{180} \cos \, x^{\circ} \, \sin \, x

Answer

π180cos⁡ x∘ cos⁡ x−sin⁡ x∘sin⁡ x\frac{\pi}{180} \cos \, x^{\circ} \, \cos \, x - \sin \, x^{\circ} \sin \, x

Explanation

Solution

It is given that sin⁡x∘ cos⁡x\sin x^{\circ}\, \cos x .
First we change x∘x^{\circ} into radian
i.e.,i.e., 1 degree =π180= \frac{\pi}{180} radian
⇒   x∘=π180x\Rightarrow \, \, \, x^{\circ} = \frac{\pi}{180} x redian
∴    sin⁡x∘.cos⁡x=sin⁡(π180x)cos⁡x\therefore \, \, \, \, \sin x^{\circ} . \cos x = \sin \left(\frac{\pi}{180} x \right) \cos x
Now differentiating w.r.t. ′x′'x', we have
⇒[cos⁡(π180x).(π180)cos⁡x]+sin⁡(π180x)(−sin⁡x)\Rightarrow \left[\cos\left(\frac{\pi}{180} x\right).\left(\frac{\pi}{180}\right) \cos x\right] +\sin\left(\frac{\pi}{180}x\right)\left(-\sin x\right)
⇒π180cos⁡x∘cos⁡x−sin⁡x∘sin⁡x\Rightarrow \frac{\pi}{180} \cos x^{\circ} \cos x -\sin x^{\circ} \sin x