Iam writing this question only to see how smart the community on is.
what is the 3rd derivative of

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.

Ifound this on google “that means that the graph of a function with direct variation has these features: it is a straight line: because the slope is the ratio y/x = k. it passes through the origin (0,0): because y = kx = k (0) = 0.” i hope this you for what your looking for