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y=cos^3(2x)+sin(3x)
We have to take the derivetive of each term separately:
Let's do cos^3(2x) first:
Let's say y_1=cos^3(2x). Note that this is cos of another function which is 2x.
Let's put u=2x. Then du=2dx
(du)/dx=2
cosu itself is another function.
Let's say z=cosu. Now, let's take the derivative:
dz/(du)=-sinu
Also, plugging cosu into the original function give us:
y_1=z^3
dy_1/dz=3z^2
The Chain Rule says:
dy_1/dx=dy_1/dz*dz/(du)*(du)/dx Let's plug them in:
dy_1/dx=(3z^2)(-sinu)(2) Now, let's substitute back for z:
dy_1/dx=(3cos^2u)(-sinu)(2) Now, let's substitute back for u:
dy_1/dx=3cos^2(2x)(-sin(2x))(2) Now, we simplify:
dy_1/dx=-6cos^2(2x)sin(2x)
Now, we do the second term:
Let's say y_2=sin(3x) But 3x is another function.
Let's say w=3x Then (dw)/dx=3 and
y_2=sinw
dy_2/(dw)=cosw
The Chain Rule says:
dy_2/dx=dy_2/(dw)*(dw)/dx Now, let's plug them in:
dy_2/dx=(cosw)(3)=3cosw Now, let's substitute back for w:
dy_2/dx=3cos(3x)
According to our own definition:
y=y_1+y_2 Then:
dy/dx=dy_1/dx+dy_2/dx
Now. we can plug the derivatives of both terms into the above equation:
dy/dx=-6cos^2(2x)sin(2x)+3cos(3x)