Calculus

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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 -

    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 -

    Dervative of cos(x) is -sin(x) so it would be negative.

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