Posted by anon 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
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 6:21am
1. 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!