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I am writing this question only to see how smart the community on Brainly is.

What is the 3rd derivative of \(f(x)=sin(x)x^3+x^2tan(x)\)

Verified Answer :

I am very confused as to why you posed this question in the way you did. People would have answered your question regardless. 

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.

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