anonymous
  • anonymous
Help please! can someone show me the steps of how to differentiate the following with respect to x 3sin^2x+sec2x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
ddxsin(ax)dx=acos(ax) use this along with chain rule for 3sin^2x
anonymous
  • anonymous
differentiate each term in turn for 3 sin^2x use the chain rule f'(x) = f'(g(x) * g'(x) also use chain rule for second term
anonymous
  • anonymous
d/dx [3sin^2(x)] = 6sin x cos x

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anonymous
  • anonymous
yup
anonymous
  • anonymous
Let y=sec 2x dy/dx=d{sec 2x}/dx =sec 2x tan2x×d(2x)/dx USing chain rule =sec 2x tan2x×(2) =2sec 2x tan2x
anonymous
  • anonymous
@jetly did you use the chain rule to get 6sin x cos x?
anonymous
  • anonymous
yes kinzan.
anonymous
  • anonymous
are you happy with the f'(x) notation or do you usually use the Leibnitz fom r kinsan : - dy/dx etc
anonymous
  • anonymous
*Leibnitz form
anonymous
  • anonymous
i use the the f'(x) notation but do not get how jetly go 2sec 2x tan2x as his final answer... arnt you meant to use a constant 'u' in the chanirule?
anonymous
  • anonymous
you can use 'u' for the 'inside function' when using the chain rule - some people find it easier to understand
anonymous
  • anonymous
i'll differentiate for you using 'u' and leibnitz notation; let y = sec2x and let u = 2x so y = secu dy/du = secu tanu du/dx = 2 so dy/dx = dy/du * du/dx = 2 secu tan u = 2 sec2x tan 2x

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