How do you differentiate #f(x)=sin(1/(2x)^3)# using the chain rule?

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Answer 1 · 2017-05-15T22:09:44

#f'(x) = -(3cos((2x)^(-3)))/(8x^4)#

The derivative of #f, (df)/dx,# can be found using the chain rule:

#(df)/dx = (df)/(dg) xx (dg)/(dx)#

Let #f(x) = sing# where #g=g(x) = 1/(2x)^3= (2x)^-3=1/8x^-3#

#(df)/(dg) = cosg = cos((2x)^-3)#

#(dg)/dx = -3/8x^-4= -3/(8x^4)#

#(df)/dx = cosg xx -3/(8x^4)#

#= -(3cos((2x^-3)))/(8x^4)#