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Differentiate y = 2 x ( x) using the product rule and chain rule. Simplify your answer.

Mathematics

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Step-by-step solution

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To differentiate y=sin⁡2xln⁡(cos⁡x)y = \sin^2 x \ln(\cos x), we will use the product rule, which states that if y=u⋅vy = u \cdot v, then dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}.

  1. Identify uu and vv.

    Let u=sin⁡2xu = \sin^2 x and v=ln⁡(cos⁡x)v = \ln(\cos x).

  2. Differentiate uu with respect to xx.

    Using the chain rule for u=(sin⁡x)2u = (\sin x)^2:

    dudx=2(sin⁡x)⋅ddx(sin⁡x)=2sin⁡xcos⁡x\frac{du}{dx} = 2(\sin x) \cdot \frac{d}{dx}(\sin x) = 2 \sin x \cos x

  3. Differentiate vv with respect to xx.

    Using the chain rule for v=ln⁡(cos⁡x)v = \ln(\cos x):

    dvdx=1cos⁡x⋅ddx(cos⁡x)=1cos⁡x⋅(−sin⁡x)=−sin⁡xcos⁡x=−tan⁡x\frac{dv}{dx} = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x) = \frac{1}{\cos x} \cdot (-\sin x) = -\frac{\sin x}{\cos x} = -\tan x

  4. Apply the product rule.

    Substitute uu, vv, dudx\frac{du}{dx}, and dvdx\frac{dv}{dx} into the product rule formula:

    dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} dydx=(sin⁡2x)(−tan⁡x)+(ln⁡(cos⁡x))(2sin⁡xcos⁡x)\frac{dy}{dx} = (\sin^2 x)(-\tan x) + (\ln(\cos x))(2 \sin x \cos x)

  5. Simplify the expression.

    dydx=−sin⁡2x(sin⁡xcos⁡x)+2sin⁡xcos⁡xln⁡(cos⁡x)\frac{dy}{dx} = -\sin^2 x \left(\frac{\sin x}{\cos x}\right) + 2 \sin x \cos x \ln(\cos x) dydx=−sin⁡3xcos⁡x+2sin⁡xcos⁡xln⁡(cos⁡x)\frac{dy}{dx} = -\frac{\sin^3 x}{\cos x} + 2 \sin x \cos x \ln(\cos x)

    We can factor out sin⁡x\sin x:

    dydx=sin⁡x(−sin⁡2xcos⁡x+2cos⁡xln⁡(cos⁡x))\frac{dy}{dx} = \sin x \left( -\frac{\sin^2 x}{\cos x} + 2 \cos x \ln(\cos x) \right)

    The final answer is sin⁡x(−sin⁡2xcos⁡x+2cos⁡xln⁡(cos⁡x))\boxed{\sin x \left( -\frac{\sin^2 x}{\cos x} + 2 \cos x \ln(\cos x) \right)}.

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