Derivative of Cos^3(x)*Sin(x) using Chain Rule

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Help!
I last did this kind of work years ago in university. Can som1 please help me with the derivative.
 
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I think the question is ((cosx)^3)*sin(x).Am I right?The formula for d(uv)/dx=u(dv/dx)+v(du/dx).And of course I think you need only formula.Do you need answer?
 
omkar13 said:
I think the question is ((cosx)^3)*sin(x).Am I right?The formula for d(uv)/dx=u(dv/dx)+v(du/dx).And of course I think you need only formula.Do you need answer?
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Thats it chain rule or something:


d(uv)/dx=u(dv/dx)+v(du/dx)

u=sin(x)
v=cos^3(x)

(dv/dx) = -3sin((x))cos^2(x)
(du/dx) = cos(x)

d(uv)/dx= -3sin^2(x)cos^2(x) + cos^4(x)

Is this correct?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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