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Miscellaneous Exercise · Q27

Q.Find the derivative of x2cos⁡π4sin⁡x\dfrac{x^2\cos\frac{\pi}{4}}{\sin x}.

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cos⁡π4=12\cos\frac{\pi}{4} = \frac{1}{\sqrt2} is a constant, so this is a constant times x2sin⁡x\frac{x^2}{\sin x}. Applying the quotient rule gives x(2sin⁡x−xcos⁡x)2 sin⁡2x\dfrac{x(2\sin x - x\cos x)}{\sqrt2\,\sin^2 x}.

The first thing to notice is that cos⁡π4=12\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt2} is just a number, not a function of xx. So the function is a constant multiple of a quotient:

y=cos⁡π4⋅x2sin⁡x=12⋅x2sin⁡xy = \cos\tfrac{\pi}{4}\cdot\frac{x^2}{\sin x} = \frac{1}{\sqrt2}\cdot\frac{x^2}{\sin x}

Step-by-step

  1. Pull the constant out front. The constant 12\dfrac{1}{\sqrt2} simply multiplies the whole derivative.

  2. Set up the quotient rule on x2sin⁡x\dfrac{x^2}{\sin x}, with u=x2u = x^2 and v=sin⁡xv = \sin x:

ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

  1. Differentiate top and bottom. u′=2xu' = 2x and v′=cos⁡xv' = \cos x.

  2. Apply the rule.

ddx(x2sin⁡x)=2xsin⁡x−x2cos⁡xsin⁡2x\frac{d}{dx}\left(\frac{x^2}{\sin x}\right) = \frac{2x\sin x - x^2\cos x}{\sin^2 x}

  1. Multiply back by the constant and factor xx from the numerator. …

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