Differentiate the function \( k(x)=5 \pi x \sin x+6 \pi x \). Write the exact answer.

To differentiate the function \( k(x) = 5 \pi x \sin x + 6 \pi x \), we can apply the product rule for the term \( 5 \pi x \sin x \) and the constant rule for the term \( 6 \pi x \). The product rule states that if you have a product of two functions \( u(x) \) and \( v(x) \), then the derivative is given by \( (uv)' = u'v + uv' \). Let: - \( u(x) = 5 \pi x \) and \( v(x) = \sin x \). Calculating the derivatives: - \( u'(x) = 5 \pi \) - \( v'(x) = \cos x \) Now, applying the product rule: \[ \frac{d}{dx}(5 \pi x \sin x) = u'v + uv' = (5 \pi)(\sin x) + (5 \pi x)(\cos x) = 5 \pi \sin x + 5 \pi x \cos x. \] Now differentiate the second term \( 6 \pi x \): \[ \frac{d}{dx}(6 \pi x) = 6 \pi. \] Putting it all together, we have: \[ k'(x) = \frac{d}{dx}(5 \pi x \sin x) + \frac{d}{dx}(6 \pi x) = (5 \pi \sin x + 5 \pi x \cos x) + 6 \pi. \] Thus, the exact answer for the derivative of the function is: \[ k'(x) = 5 \pi \sin x + 5 \pi x \cos x + 6 \pi. \]