Use the rule for differentiating products of functions: ()' signifies derivative
(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'
But
Result is
(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)
You could also have cast your formula in the form
sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]
then calculated the derivative of
29/2*[cos(X)-cos(3X)]
which is
29/2*[-si(X) +3*sin(3X)]
The challenge for you is to prove that the two forms are equivalent
29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]
Question 1: (i) Find the complex conjugate of (3+5i)/(1+2i)
(ii) Differentiate (sin x)x w.r.t. x.
(iii) Find all the seventh roots of (3+4i).
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