How do you differentiate #f(x) = (x^3+cos3x)^(1/2)?# using the chain rule?

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Answer 1 · 2018-04-07T08:41:02

#f'(x)=3/2(x^2-sin3x)(x^3+cos3x)^(-1/2)#

#(dy)/(dx)=(dy)/(du)color(red)((du)/(dx))#

#y=f(x)=(x^3+cos3x)^(1/2)#

#color(red)(u=x^3+cos3x=>(du)/(dx)=3x^2-3sin3x)#

#y=u^(1/2)=>(dy)/(du)=1/2u^(-1/2)#

#:.f'(x)=(dy)/(dx)=1/2u^(-1/2)xxcolor(red)(3x^2-3sin3x)#

substituting and tiding up

#f'(x)=3/2(x^2-sin3x)(x^3+cos3x)^(-1/2)#