Question Statement

Find the derivative of the function:

$y=(1+\cos x)(x-\sin x)$

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Background and Explanation

This problem requires the product rule for differentiation, which is used when differentiating the product of two functions. You should also be familiar with basic trigonometric derivatives ($\frac{d}{dx}\cos x=-\sin x$) and the Pythagorean identity ${\sin }^{2}x+{\cos }^{2}x=1$.

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Solution

We are given: $y=(1+\cos x)(x-\sin x)$

To differentiate this function, we recognize it as a product of two functions:

• Let $u=1+\cos x$
• Let $v=x-\sin x$

Step 1: Apply the product rule $\frac{d}{dx}(uv)=\frac{du}{dx}\cdot v+u\cdot \frac{dv}{dx}$

$\frac{dy}{dx}=\frac{d}{dx}(1+\cos x)\cdot (x-\sin x)+(1+\cos x)\cdot \frac{d}{dx}(x-\sin x)$

Step 2: Compute the derivatives of the individual terms

• $\frac{d}{dx}(1+\cos x)=0-\sin x=-\sin x$
• $\frac{d}{dx}(x-\sin x)=1-\cos x$

Substituting these back:

$\frac{dy}{dx}=(-\sin x)(x-\sin x)+(1+\cos x)(1-\cos x)$

Step 3: Expand both products

First term: $(-\sin x)(x-\sin x)=-x\sin x+{\sin }^{2}x$

Second term: $(1+\cos x)(1-\cos x)=1-{\cos }^{2}x$ (this is a difference of squares: $(1+\cos x)(1-\cos x)=1^2-{\cos }^{2}x$)

So: $\frac{dy}{dx}=-x\sin x+{\sin }^{2}x+1-{\cos }^{2}x$

Step 4: Simplify using the Pythagorean identity

Recall that $1-{\cos }^{2}x={\sin }^{2}x$. Substituting this:

$\frac{dy}{dx}=-x\sin x+{\sin }^{2}x+{\sin }^{2}x$

Step 5: Combine like terms

$\frac{dy}{dx}=-x\sin x+2{\sin }^{2}x$

Or rearranged: $\frac{dy}{dx}=2{\sin }^{2}x-x\sin x$

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Key Formulas or Methods Used

• Product Rule: $\frac{d}{dx}[u(x)v(x)]=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$
• Derivative of cosine: $\frac{d}{dx}(\cos x)=-\sin x$
• Derivative of sine: $\frac{d}{dx}(\sin x)=\cos x$
• Pythagorean Identity: $1-{\cos }^{2}x={\sin }^{2}x$
• Algebraic expansion: $(a+b)(a-b)=a^2-b^2$

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Summary of Steps

1. Identify the two factors in the product: $(1+\cos x)$ and $(x-\sin x)$
2. Apply the product rule: $\frac{dy}{dx}=u^{\prime }v+u{v}^{\prime }$
3. Differentiate each factor: $\frac{d}{dx}(1+\cos x)=-\sin x$ and $\frac{d}{dx}(x-\sin x)=1-\cos x$
4. Expand the resulting expression: $(-\sin x)(x-\sin x)+(1+\cos x)(1-\cos x)$
5. Simplify using the identity $1-{\cos }^{2}x={\sin }^{2}x$ to combine trigonometric terms
6. Combine like terms to obtain the final answer: $2{\sin }^{2}x-x\sin x$