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Miscellaneous Examples · Example 43

Q.Differentiate sin⁡2x\sin^2 x w.r.t. ecos⁡xe^{\cos x}.

Yanam CbseNCERTSubjective· 3mImportance★★★★★
Appeared in past exams:MHT-CET 2024· Set pcm-2024-05-03-E· 2mexact
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To differentiate sin⁡2x\sin^2 x with respect to ecos⁡xe^{\cos x}, we use the chain rule in parametric form: dydu=dy/dxdu/dx\frac{dy}{du} = \frac{dy/dx}{du/dx}. The final result is −2cos⁡xecos⁡x\boxed{-\frac{2\cos x}{e^{\cos x}}}.

The question asks us to differentiate one function with respect to another — not the usual dy/dxdy/dx where both are functions of xx. This is a classic "parametric differentiation" problem. Think of it this way: both sin⁡2x\sin^2 x and ecos⁡xe^{\cos x} are expressed in terms of a common parameter xx. So we can find the derivative of the first with respect to the second by taking the ratio of their individual derivatives with respect to xx.

The chain rule is the backbone here. If y=sin⁡2xy = \sin^2 x and u=ecos⁡xu = e^{\cos x}, then:

dydu=dy/dxdu/dx\frac{dy}{du} = \frac{dy/dx}{du/dx}

provided du/dx≠0du/dx \neq 0. This is simply the derivative of yy with respect to xx divided by the derivative of uu with respect to xx.

Let's work through it step by step.

  1. Set up the functions.

    Let y=sin⁡2xy = \sin^2 x and u=ecos⁡xu = e^{\cos x}. We need dydu\frac{dy}{du}.

  2. Differentiate y=sin⁡2xy = \sin^2 x with respect to xx.

    This is a composition: (sin⁡x)2(\sin x)^2. Using the chain rule:

dydx=2sin⁡x⋅ddx(sin⁡x)=2sin⁡xcos⁡x.\frac{dy}{dx} = 2 \sin x \cdot \frac{d}{dx}(\sin x) = 2 \sin x \cos x.

Alternatively, you can write 2sin⁡xcos⁡x=sin⁡2x2 \sin x \cos x = \sin 2x, but we'll keep it as 2sin⁡xcos⁡x2 \sin x \cos x for now.

  1. Differentiate u=ecos⁡xu = e^{\cos x} with respect to xx. Again, chain rule: derivative of esomethinge^{\text{something}} is esomethinge^{\text{something}} times the derivative of that something.

dudx=ecos⁡x⋅ddx(cos⁡x)=ecos⁡x⋅(−sin⁡x)=−ecos⁡xsin⁡x.\frac{du}{dx} = e^{\cos x} \cdot \frac{d}{dx}(\cos x) = e^{\cos x} \cdot (-\sin x) = -e^{\cos x} \sin x.

  1. Form the ratio.

dydu=dy/dxdu/dx=2sin⁡xcos⁡x−ecos⁡xsin⁡x.\frac{dy}{du} = \frac{dy/dx}{du/dx} = \frac{2 \sin x \cos x}{-e^{\cos x} \sin x}.

  1. Simplify. …

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