^{2}sin(2x) at x=-π

^{2}cosx

1. Find the slope of the function y=2cosxsin2x at x=π/2

2. Find the equation of the line that is tangent to y=x^{2}sin(2x) at x=-π

3. Determine the second derivative for y=x^{2}cosx

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y ’(x) = 2 (-sin(x)) sin(2x) + 4 cos(x) cos(2x)

y ’(pi/2) = 2(-1)(0) + 4(0)(-1) = 0

2. Find the equation of the line that is tangent to y = x^2 sin(2x) at x=-π

y = x^2 sin(2x)

y ’(x) = 2x sin(2x) + 2 x^2 cos(2x)

y ’(-pi) = -2pi(0) + 2pi^2 (1) = 2 pi^2

y(-pi) = pi^2 (0) = 0

Tangent line:

y - 0 = 2 pi^2 (x - -pi)

y = 2 pi^2 x + 2 pi^3

3. Determine the second derivative for y = x^2 cos(x)

y ’ = 2x cos(x) - x^2 sin(x)

y ’’ = 2 cos(x) - 2x sin(x) - 2x sin(x) - x^2 cos(x)

y ’’ = (2 - x^2) cos(x) - 4x sin(x)

1. Find the slope of the function y(x) = 2 cos(x) sin(2x) at x=π/2

y'(x)=-2sin(x)sin(2x)+4cos(x)cos(2x), which at x=π/2 is 0

since sin(π)=0 and cos(π)=0

2. Find the equation of the line that is tangent to y=x^{2}sin(2x) at x=-π

y'(x)=2xsin(2x)+2x^{2}cos(2x)

y(-π)=0, y'(-π)=-2πsin(-4π)+2π^{2}cos(-4π)=2π^{2}

The line has the equation y=2π^{2}x+b, which, for a line through the point (-π,0), of that slope

gives b=2π^{3}, so have y=2π^{2}x+2π^{3}

3.Determine the second derivative for y=x^{2}cosx

y'(x)=2xcosx-x^{2}sinx

y''(x)=2cosx-2xsinx-2xsinx-x^{2}cosx=(2-x^{2})cosx-4xsinx

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