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Miscellaneous Exercise · Q16

Q.Find the derivative of cos⁡x1+sin⁡x\dfrac{\cos x}{1 + \sin x}.

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The derivative of cos⁡x1+sin⁡x\frac{\cos x}{1 + \sin x} is found using the Quotient Rule, simplifying via cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1, and the final result is −11+sin⁡x-\frac{1}{1 + \sin x}.

The Quotient Rule is the natural choice here because we have one function divided by another. The rule says: for y=uvy = \frac{u}{v}, the derivative is dydx=v⋅u′−u⋅v′v2\frac{dy}{dx} = \frac{v \cdot u' - u \cdot v'}{v^2}. The trick is to avoid messy algebra by spotting simplifications early — especially the identity cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1.

Let’s work through it step by step.

  1. Identify uu and vv

    Let u=cos⁡xu = \cos x and v=1+sin⁡xv = 1 + \sin x.

    Then u′=−sin⁡xu' = -\sin x and v′=cos⁡xv' = \cos x.

  2. Apply the Quotient Rule

dydx=(1+sin⁡x)(−sin⁡x)−(cos⁡x)(cos⁡x)(1+sin⁡x)2\frac{dy}{dx} = \frac{(1 + \sin x)(-\sin x) - (\cos x)(\cos x)}{(1 + \sin x)^2}

  1. Simplify the numerator Expand the first term: (1+sin⁡x)(−sin⁡x)=−sin⁡x−sin⁡2x(1 + \sin x)(-\sin x) = -\sin x - \sin^2 x. Subtract the second term: −cos⁡2x-\cos^2 x. So numerator becomes:

−sin⁡x−sin⁡2x−cos⁡2x-\sin x - \sin^2 x - \cos^2 x

  1. Use the Pythagorean identity Recall sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. So −sin⁡2x−cos⁡2x=−(sin⁡2x+cos⁡2x)=−1-\sin^2 x - \cos^2 x = -(\sin^2 x + \cos^2 x) = -1. Thus the numerator simplifies to:

−sin⁡x−1-\sin x - 1

  1. Write the derivative

dydx=−sin⁡x−1(1+sin⁡x)2\frac{dy}{dx} = \frac{-\sin x - 1}{(1 + \sin x)^2}

  1. Factor and cancel Factor −1-1 from the numerator: −(sin⁡x+1)-\left(\sin x + 1\right). Notice sin⁡x+1=1+sin⁡x\sin x + 1 = 1 + \sin x, so:

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