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Exercise 5.7 · Q7

Q.Find dydx\frac{dy}{dx} in the following: e6xcos⁡3xe^{6x} \cos 3x

Puducherry CbseNCERTSubjective· 2mImportance★★★★★
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This problem uses the Product Rule combined with the Chain Rule — treat e6xe^{6x} and cos⁡3x\cos 3x as two separate functions, differentiate each, then add the cross-terms. The final derivative is dydx=e6x(6cos⁡3x−3sin⁡3x)\frac{dy}{dx} = e^{6x}(6\cos 3x - 3\sin 3x).

We are given y=e6xcos⁡3xy = e^{6x} \cos 3x. This is a product of two functions: u=e6xu = e^{6x} and v=cos⁡3xv = \cos 3x. To differentiate a product, we use the Product Rule: ddx(uv)=u′v+uv′\frac{d}{dx}(uv) = u'v + uv'.

But each of these functions is itself a composition — e6xe^{6x} is ee raised to 6x6x, and cos⁡3x\cos 3x is cos⁡\cos of 3x3x. So each requires the Chain Rule as well. Let’s break it down step by step.

  1. Differentiate u=e6xu = e^{6x}. The derivative of esomethinge^{\text{something}} is esomethinge^{\text{something}} times the derivative of that something. Here, the "something" is 6x6x, whose derivative is 66. So:

u′=e6x⋅6=6e6x.u' = e^{6x} \cdot 6 = 6e^{6x}.

  1. Differentiate v=cos⁡3xv = \cos 3x. The derivative of cos⁡(something)\cos(\text{something}) is −sin⁡(something)-\sin(\text{something}) times the derivative of that something. Here, the "something" is 3x3x, whose derivative is 33. So:

v′=−sin⁡3x⋅3=−3sin⁡3x.v' = -\sin 3x \cdot 3 = -3\sin 3x.

  1. Apply the Product Rule. The Product Rule says:

dydx=u′v+uv′.\frac{dy}{dx} = u'v + uv'.

Substitute uu, u′u', vv, and v′v':

dydx=(6e6x)(cos⁡3x)+(e6x)(−3sin⁡3x).\frac{dy}{dx} = (6e^{6x})(\cos 3x) + (e^{6x})(-3\sin 3x).

  1. Simplify. Factor out the common factor e6xe^{6x}: …

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