# How do you use part 1 of Fundamental Theorem of Calculus to find the derivative of the function #y=int (6+v^2)^10 dv# from sinx to cosx?

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To find the derivative of the function ( y = \int_{\sin x}^{\cos x} (6+v^2)^{10} , dv ) using Part 1 of the Fundamental Theorem of Calculus, we first need to evaluate the integrand at the upper limit (cos x) and the lower limit (sin x). Then, we differentiate the result with respect to x. The derivative of the function with respect to x is given by:

[ \frac{d}{dx} \left[ \int_{\sin x}^{\cos x} (6+v^2)^{10} , dv \right] = (6 + \cos^2 x)^{10} \cdot \frac{d}{dx} \cos x - (6 + \sin^2 x)^{10} \cdot \frac{d}{dx} \sin x ]

[ = (6 + \cos^2 x)^{10} \cdot (-\sin x) - (6 + \sin^2 x)^{10} \cdot \cos x ]

[ = -\sin x \cdot (6 + \cos^2 x)^{10} - \cos x \cdot (6 + \sin^2 x)^{10} ]

This is the derivative of the given function with respect to x.

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