1. messy, but correct. I'd have gone on to
(6x^2 + 2x + 3)e^(x^2)
2. almost. Why the "-" sign? d(sinx)/dx = +cosx
e^(sinx) (1/x + lnx cosx)
4. brute force, using the quotient rule:
((2cos 2x)(cosx) - sin 2x (-sinx))/cosx)^2
= (2 cosx cos2x + sinx sin2x)/(cosx)^2
messy. How about simplifying first?
sin2x/cosx = 2sinx cosx / cosx = 2sinx
f' = 2cosx
I'll let you convince yourself that the two are equal
5. Hmmm. I get
f = √x/(3x+1)
f' = ((1/2√x)(3x+1) - √x(3))/(3x+1)^2
= ((3x+1) - 3√x 2√x)/(2√x (3x+1)^2)
= (1-3x)/(2√x (3x+1)^2)
a little algebra mixup in the top?
6. If you mean
f = sin^4(3x+1)-sin(3x+1), just use the good old chain rule on both terms:
f' = 4sin^3(3x+1)(cos(3x+1))(3) - cos(3x+1)(3)
= 3cos(3x+1)(4sin^3(3x+1) - 1)
7. where'd that 2 come from?
1 + y' = -sin(xy)(y+xy')
1 + y' = -ysin(xy) - xsin(xy) y'
y' = -(1+y sin(xy))/(1+x sin(xy))
Dervative of cos(x) is -sin(x) so it would be negative.
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