How do you differentiate #f(x)=cos(x^3)#?
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#d/(dx)cos(x^3)=-3x^2sin(x^3)#
Use chain rule: #(dy)/(dx)=(dy)/(du)*(du)/(dx)#
#y=cos(x^3)#, let #u=x^3#
Then #(du)/(dx)=3x^2# and #(dy)/(du)=-sinu=-sin(x^3)#
So #(dy)/(dx)=3x^2*-sin(x^3)=-3x^2sin(x^3)#
The answer is #-3x^2 sin(x^3)#
I mainly use formulas because some of them are easy to memorize and they help you see the answer right away, but you can also use the "u substitution." I think that's what is officially known as the "Chain Rule"
#color (red) (d/dx cos x = (cosx)'=-(x)'sinx=-sinx)# and when it's not #x# but any other variable, like #5x# for example, the formula is #color (red) (d/(du) cos u =(cos u)'= -(u)'sinu=-u'sinu)#
Note that #color (red) (u')# is the derivative of #color (red) u#
Our problem #f(x)=cos(x^3)#
Since it's not simply #x# but #x^3#, the first formula will not work but the second will.
#f'(x)=(cos(x^3))'=-3x^2 sin(x^3)#
Another method: "u substitution"
#f(x)=cos(x^3)#
Let's say #u=x^3 => f(u)=cosu#
#f'(u)=-u'sinu#
And the derivative of #u=(u)'=(x^3)'=3x^2#
#=>f'(u)=-3x^2(sin(u))#
Substitute back #u=x^3#
#f'(x)=-3x^2(sin(x^3))=-3x^2sin(x^3)#
Hope this helps :)
