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+1-3squarerootofx/(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)

= (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))

- 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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