Let fx6sinx4sinx2cosx Then fx The equation of the tangent li

Let f(x)=6sinx/4sinx+2cosx. Then f\'(x)= The equation of the tangent line to y = f(x) at a = 0 can be written in the form y = mx + b where m = and b =

Solution

d/dx((6 sin(x))/(4 sin(x)+2 cos(x))) | Factor out constants: = | 6 (d/dx((sin(x))/(4 sin(x)+2 cos(x)))) | Use the quotient rule, d/dx(u/v) = (v ( du)/( dx)-u ( dv)/( dx))/v^2, where u = sin(x) and v = 4 sin(x)+2 cos(x): = | 6 ((4 sin(x)+2 cos(x)) (d/dx(sin(x)))-sin(x) (d/dx(4 sin(x)+2 cos(x))))/(4 sin(x)+2 cos(x))^2 | Differentiate the sum term by term and factor out constants: = | (6 ((4 sin(x)+2 cos(x)) (d/dx(sin(x)))-sin(x) (4 (d/dx(sin(x)))+2 (d/dx(cos(x))))))/(4 sin(x)+2 cos(x))^2 | The derivative of cos(x) is -sin(x): = | (6 ((4 sin(x)+2 cos(x)) (d/dx(sin(x)))-sin(x) (4 (d/dx(sin(x)))+2 (-sin(x)))))/(4 sin(x)+2 cos(x))^2 | The derivative of sin(x) is cos(x): = | (6 ((4 sin(x)+2 cos(x)) (d/dx(sin(x)))-sin(x) (4 cos(x)-2 sin(x))))/(4 sin(x)+2 cos(x))^2 | The derivative of sin(x) is cos(x): = | (6 (cos(x) (4 sin(x)+2 cos(x))-sin(x) (4 cos(x)-2 sin(x))))/(4 sin(x)+2 cos(x))^2 The tangent line at a = 0, is y = 3x, so the slope (m) is 3 and your b is 0.