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Worked solution

Use a derivative identity for an even exponential composition

Given y=ecos⁡xy=e^{\cos x}, show y′′=−ycos⁡x−y′sin⁡xy^{\prime\prime}=-y\cos x-y^{\prime}\sin x. Show that the first two nonzero terms in its expansion are e−12ex2e-\tfrac12ex^2. marks

CJC · Review problems · Question 4

Answer

y′′=−ycos⁡x−y′sin⁡xy^{\prime\prime}=-y\cos x-y^{\prime}\sin x; ecos⁡x=e−ex2/2+O(x4)e^{\cos x}=e-ex^2/2+O(x^4).

Differentiate the composition and its product

y′=−ecos⁡xsin⁡x=−ysin⁡xy^{\prime}=-e^{\cos x}\sin x=-y\sin x. Differentiate again: y′′=−y′sin⁡x−ycos⁡xy^{\prime\prime}=-y^{\prime}\sin x-y\cos x, as required.

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Evaluate at the expansion point

y(0)=ey(0)=e, y′(0)=0y^{\prime}(0)=0 and y′′(0)=−ey^{\prime\prime}(0)=-e. Therefore ecos⁡x=e−ex2/2+O(x4)e^{\cos x}=e-ex^2/2+O(x^4). The function is even, so there is no cubic term; these are its first two nonzero terms.

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Common mistakes

The chain rule supplies the minus sign in the first derivative.

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