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

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

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