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January 27, 2015

January 27, 2015

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

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

A.be^bx^2

B.bxe^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

A.du/dx=2x+2

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

C.dv/dx=4xe^3x

D.dv/dx=4xe^4x

- Calculus 12th grade (double check my work please) -
**MathMate**, Sunday, March 6, 2011 at 6:21am1. You need to follow the chain rule. 2cosh(2x) is only the second part:

d(ln(sinh(2x)/dx

=d(ln(sinh(2x))/d(sinh(2x)) * d(sihn(2x))/dx

=1/sinh(2x) * 2cosh(2x)

=2cosh(2x)/sinh(2x)

=2coth(2x)

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

correct.

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

correct.

3. C is correct

4. C is correct

5. B is correct.

Keep up the good work!

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