Find the derivative of f(x). {:[f(x)=x sin(x)],[f^(')(x)=]:}

Apply Product Rule: Apply the product rule for differentiation. The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. So, for f(x) = x sin(x), we have: f'(x) = d/dx(x) * sin(x) + x * d/dx(sin(x))Differentiate Each Part: Differentiate each part. 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: f'(x) = 1 * sin(x) + x * cos(x)Simplify Expression: Simplify the expression. f'(x) = sin(x) + x cos(x) This is the simplified form of the derivative of f(x) = x sin(x).

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Find derivatives of sine and cosine functions

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Q. Find the derivative of

f(x)

\newline

\begin{array}{l} f(x)=x \sin (x) \\ f^{\prime}(x)= \end{array}

Apply Product Rule: Apply the product rule for differentiation. $\newline$ The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. $\newline$ So, for $f(x) = x \sin(x)$ , we have: $\newline$ $f'(x) = \frac{d}{dx}(x) \cdot \sin(x) + x \cdot \frac{d}{dx}(\sin(x))$
Differentiate Each Part: Differentiate each part. $\newline$ The derivative of $x$ with respect to $x$ is $1$ , and the derivative of $\sin(x)$ with respect to $x$ is $\cos(x)$ . $\newline$ So we have: $\newline$ $f'(x) = 1 \cdot \sin(x) + x \cdot \cos(x)$
Simplify Expression: Simplify the expression. $f'(x) = \sin(x) + x \cos(x)$ This is the simplified form of the derivative of $f(x) = x \sin(x)$ .