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anonymous
 5 years ago
Help please!
can someone show me the steps of how to differentiate the following with respect to x
3sin^2x+sec2x
anonymous
 5 years ago
Help please! can someone show me the steps of how to differentiate the following with respect to x 3sin^2x+sec2x

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ddxsin(ax)dx=acos(ax) use this along with chain rule for 3sin^2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0differentiate 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
 5 years ago
Best ResponseYou've already chosen the best response.0d/dx [3sin^2(x)] = 6sin x cos x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let 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
 5 years ago
Best ResponseYou've already chosen the best response.0@jetly did you use the chain rule to get 6sin x cos x?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you happy with the f'(x) notation or do you usually use the Leibnitz fom r kinsan :  dy/dx etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0you can use 'u' for the 'inside function' when using the chain rule  some people find it easier to understand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'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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