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Lilbossman2203
30.08.2019 •
Mathematics
Iam writing this question only to see how smart the community on is.
what is the 3rd derivative of
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Ответ:
First derivative:
f'(x) = 3x^2sin(x) + x^3cos(x) + 2xtan(x) + x^2sec^2(x)
Second derivative:
f''(x) = 6xsin(x) + 3x^2cos(X) + 3x^2cos(x) - x^3sin(x) + 2tan(x) + 2xsec^2(x) + 2xsec^2(x) + 2x^2sec^2(x)tan(x)
f''(x) = 6xsin(x) + 6x^2cos(x) - x^3sin(x) + 2tan(x) + 4xsec^2(x) +
Third Derivative:
f'''(x) = 6sin(x) + 6xcos(x) + 12xcos(x) - 6x^2sin(x) - 3x^2sin(x) - x^3cos(x) + 2sec^2(x) + 8xsec^2(x)tan(x) + 4sec^2(x) + 4x^2sec^2(x)tan^2(x) + 4xsec^2(x)tan(x) + 2x^2sec^4(x)
f'''(x) = 6sin(x) + 18xcos(x) - 9x^2sin(x) - x^3cos(x) + 12xsec^2(x)tan(x) + 6sec^2(x) + 4x^2sec^2(x)tan^2(x) + 2x^2sec^4(x)
f'''(x) = (6 - 9x^2)sin(x) + (18x - x^3)cos(x) + 12xsec^2(x)tan(x) + 6sec^2(x) + 4x^2sec^2(x)tan^2(x) + 2x^2sec^4(x)
There are many more ways this can be factored, but if this is just to test whether someone knows the basics of calculus, this should be sufficient.
Ответ:
Lol is this serious or a joke