Answer:
x=0,y = -2
Step-by-step explanation:
hope i helped
:)
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:
<span>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)
</span>f''(x) = 6xsin(x) + 6<span>x^2cos(x) </span>- 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) + <span>4<span>x^2</span><span>sec^2(</span>x)<span>tan^2(</span>x) + 4x<span>sec^2(</span>x)tan(x) + 2x^2<span>sec^4(</span>x)
</span>f'''(x) = 6sin(x) + 18xcos(x) - 9x^2sin(x) - x^3cos(x) + 12xsec^2(x)tan(x) + 6sec^2(x) + <span>4x^2sec^2(x)tan^2(x) + 2x^2sec^4(x)
</span>
f'''(x) = (6 - 9x^2)sin(x) + (18x - <span>x^3)</span>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.
Answer:
-1
Step-by-step explanation:
Answer:
2
Step-by-step explanation:
0! + 1!
= !1 / 1 + 1!
= 1 + 1
= 2