Posted by Anonymous on Monday, October 15, 2012 at 12:04pm.
Find the dervatives:
1. f(x)=(3x+1)e^x^2
2. y = e^(sin x) ln(x)
3. f(x)=(x^(2)+x)^23
4. f(x)=sin(2x)/cosx
5. square root of x/(3x+1)
6. f(x)=sin^4(3x=1)sin(3x+1)
7. x+y=cos(xy)
The answers I got:
1. 6x^(2)e^(x^2)+2xe^(x^2)+3e^(x^2)
2. e^sinx/x + lnxe^(sinx)cosx
3. 46x+23(x^2+x)^22
4. ?
5. 3x+13squarerootofx/(2squarerootofx)(3x+1)^w
6. ?
7. (ysin(xy)1)/(2+xsin(xy))

Calculus  Steve, Monday, October 15, 2012 at 2:34pm
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)
3. correct.
23(2x+1)(x^2+x)^22
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)
= (13x)/(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))

Calculus  Bill, Monday, October 15, 2012 at 10:16pm
Dervative of cos(x) is sin(x) so it would be negative.
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