Answer:
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Step-by-step explanation:
If you look at the graph of sin(n), you can notice that it oscillates every kpi/2 when k is odd. This oscillation proves that sin(n) diverges.
For #32, the answer is the third one because you mixed up your x and y values, you were on the right track though
#33 is the last one
The picture is not clear. let me assume
y = (x^4)ln(x^3)
product rule :
d f(x)g(x) = f(x) dg(x) + g(x) df(x)
dy/dx = (x^4)d[ln(x^3)/dx] + d[(x^4)/dx] ln(x^3)
= (x^4)d[ln(x^3)/dx] + 4(x^3) ln(x^3)
look at d[ln(x^3)/dx]
d[ln(x^3)/dx]
= d[ln(x^3)/dx][d(x^3)/d(x^3)]
= d[ln(x^3)/d(x^3)][d(x^3)/dx]
= [1/(x^3)][3x^2] = 3/x
... chain rule (in detail)
end up with
dy/dx = (x^4)[3/x] + 4(x^3) ln(x^3)
= x^3[3 + 4ln(x^3)]
