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Posted by on Sunday, March 6, 2011 at 12:38am.

1.)Find dy/dx when y= Ln (sinh 2x)
my answer >> 2coth 2x.

2.)Find dy/dx when sinh 3y=cos 2x
A.-2 sin 2x
B.-2 sin 2x / sinh 3y
C.-2/3tan (2x/3y)
D.-2sin2x / 3 cosh 3yz...>> my answer.

2).Find the derivative of y=cos(x^2) with respect to x.

A.-sin (2x)
B.-2x sin (x^2)<<<< my answer.
C.-sin (2x) cos (x^2)
D.-2x sin (x^2) cos (x^2)

3).Find the derivative of y=sin^2(4x) cos (3x) with respect to x.

A. 8 sin (4x) cos (3x) - 3 sin^2 (4x) sin (3x)
B. 8 cos (4x) cos (3x) - 3 sin^2 (4x) sin (3x)
C. 8 sin (4x) cos (4x) cos (3x) - 3 sin^2 (4x) sin (3x)<<<<<my choice .not sure though
D. 8 sin (4x) cos (4x) cos (3x) - 3 sin^2 (4x) sin (3x) cos (3x)

4).Find the derivative of y=e^bx2 with respect to x^bx^2
C.2 bxe^bx^2 <<<<< my choice
D.bx^2 e^bx^2

5).Calculate dy/dx when y=(x^2+2)e^4x as the derivative of a product,letting u=x^2+2 and v=e^4x. Which of the following is a step in your solution

B.du/dx=2x <<< my choice

  • Calculus 12th grade (double check my work please) - , Sunday, March 6, 2011 at 6:21am

    1. You need to follow the chain rule. 2cosh(2x) is only the second part:
    =d(ln(sinh(2x))/d(sinh(2x)) * d(sihn(2x))/dx
    =1/sinh(2x) * 2cosh(2x)

    2. "D.-2sin2x / 3 cosh 3yz...>> my answer. "

    2. "B.-2x sin (x^2)<<<< my answer. "

    3. C is correct

    4. C is correct

    5. B is correct.

    Keep up the good work!

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