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Chapter 3: Problem 28

Find \(d y / d x\) for the following functions. $$y=\frac{x \sin x}{1+\cos x}$$

Short Answer

Expert verified
Question: Find the derivative of the following function with respect to x: $$y = \frac{x\sin x}{1 + \cos x}$$. Answer: The derivative of the given function with respect to x is: $$\frac{dy}{dx} = \frac{\sin x + x\cos x(1+\cos x) + x\sin^2 x}{(1+\cos x)^2}$$

Step by step solution

01

Find the derivative of u with respect to x.

To find the derivative of the function u, we have to apply the product rule for derivatives. The product rule states that for a function u = fg, where f and g are functions of x, the derivative of u with respect to x is: $$\frac{du}{dx} = f'g + fg'$$ In our case, $$f = x$$ and $$g = \sin x$$. We know that the derivative of x with respect to x is 1 and the derivative of sin x with respect to x is cos x. So, we have: $$\frac{du}{dx} = f'g + fg' = (1)(\sin x) + (x)(\cos x) = \sin x + x\cos x$$
02

Find the derivative of v with respect to x.

We need to find the derivative of the function v = 1 + cos x with respect to x. We know that the derivative of a constant is zero, and the derivative of cos x with respect to x is -sin x. Therefore, we have: $$\frac{dv}{dx} = \frac{d}{dx}(1 + \cos x) = \frac{d}{dx}(1) + \frac{d}{dx}(\cos x) = 0 - \sin x = - \sin x$$
03

Apply the quotient rule.

Now we can apply the quotient rule using the derivatives we found in steps 1 and 2: $$\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{(\sin x + x\cos x)(1 + \cos x) - (x\sin x)(-\sin x)}{(1+\cos x)^2}$$
04

Simplify the expression.

Let's simplify the numerator: $$\frac{dy}{dx} = \frac{(\sin x + x\cos x)(1+\cos x) + x\sin^2 x}{(1+\cos x)^2}$$ Distribute and combine like terms: $$\frac{dy}{dx} = \frac{\sin x + \sin x\cos x + x\cos x + x\cos^2 x + x\sin^2 x}{(1+\cos x)^2}$$ After combining the terms, we have our final simplified expression for the derivative: $$\frac{dy}{dx} = \frac{\sin x + x\cos x(1+\cos x) + x\sin^2 x}{(1+\cos x)^2}$$

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