Forgot, what is the anti-derivative and derivative for sin(x^2)
some places even teach it in 9th
I’m in 9th bruh
This is kind of contradicting but if they just started (started from rest so at the very beginning 0m/s) 18.345 degrees
If they started from an initial velocity of 28 per second I don’t want to solve that
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the derivative is 2xcos(x^2). Idk about the anti-derivative
If the derivative is 2xcosx then it’s probably -(1/2)xcos(x^2) for the anti-derivative
I mostly needed one to know the other anyways, thanks assuming this is right.
That won’t work, but I see what you’re going for. If you use chain rule on the cos(x^2), a 2x will come out, and you want it it to cancel out with the 1/2x, but it won’t work that way, because you have to use product rule
Ah ok, so it would be 1/(2x) then? Since then the 2x would cancel out the 1/(2x) as it goes from -cos to sin
no, if you multiply it by anything that’s not constant, you have to use product rule, so the derivative of -1/(2x)cos(x^2) would be 1/(2x^2)cos(x^2) + sin(x^2)
Ah ok
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Derivative is 2xcos(x^2)
Derivative for sin(u) is cos(u) and derivative for x^2 is 2x. Due to chain rule you multiply them together.
The anti-derivative seems like an integration by parts problem, so just give me a bit to solve that one.
Nope it is not an integration by parts problem (the power of the x spawned by the integral would diverge to infinity). As per this video it appears it requires a Taylor series approximation to calculate.
Oh hey baiting works
change of base
log(7)/log(9x) = -log(7)/log(4x+1)
cross multiply
log(7) * log(4x+1) = -log(7) * log(9x)
divide both sides by log(7)
log(4x+1) = -log(9x)
property of negative log
log(4x+1) = 1/log(9x)
property of logs being 1 to 1 (aka bye bye logs)
4x+1 = 1/9x
36x^2 + 9x - 1 = 0
(12x - 1)(3x+1) = 0
x = 1/12, -1/3
-1/3 doesnt work because it makes log9x(7) explode
so x = 1/12
(i totally didnt look at the answer sheet)