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

too much work

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!

Let's go through each question one by one and determine the correct answer step by step.

1) Find dy/dx when y = ln(sinh(2x))

To find the derivative, we will use the chain rule. The derivative of ln(u) is 1/u multiplied by the derivative of u with respect to x.

Let u = sinh(2x)

Now, we need to find the derivative of u with respect to x, which is du/dx.

Using the chain rule, du/dx = cosh(2x) * 2

Next, applying the chain rule to the original equation, dy/dx = (1/u) * du/dx

Substituting the value of u and du/dx, we have dy/dx = (1/sinh(2x)) * (cosh(2x) * 2)

Simplifying, dy/dx = 2cosh(2x)/sinh(2x)

Therefore, the correct answer is dy/dx = 2cosh(2x)/sinh(2x).

2) Find dy/dx when sinh(3y) = cos(2x)

To find the derivative, we will use implicit differentiation.

Differentiate both sides of the equation with respect to x:

d/dx(sinh(3y)) = d/dx(cos(2x))

Using the chain rule on the left side, we have:

3cosh(3y) * dy/dx = -sin(2x) * 2

Dividing both sides by 3cosh(3y), we have:

dy/dx = -2sin(2x) / 3cosh(3y)

Therefore, the correct answer is dy/dx = -2sin(2x) / 3cosh(3y).

3) Find the derivative of y = sin^2(4x) * cos(3x)

To find the derivative, we will use the product rule.

Let u = sin^2(4x) and v = cos(3x).

Now, we can find du/dx by applying the chain rule:

du/dx = 2sin(4x) * cos(4x) * 4

And dv/dx = -sin(3x) * 3

Now, using the product rule, the derivative of y with respect to x is given by:

dy/dx = du/dx * v + u * dv/dx

Substituting the values of du/dx, v, and dv/dx, we have:

dy/dx = 8sin(4x) * cos(4x) * cos(3x) + sin^2(4x) * (-sin(3x) * 3)

Simplifying, dy/dx = 8sin(4x) * cos(4x) * cos(3x) - 3sin^2(4x) * sin(3x)

Therefore, the correct answer is dy/dx = 8sin(4x) * cos(4x) * cos(3x) - 3sin^2(4x) * sin(3x).

4) Find the derivative of y = e^(bx^2)

To find the derivative, we will use the chain rule.

Let u = bx^2.

Now, we can find du/dx by differentiating u with respect to x:

du/dx = 2bx

Now, using the chain rule, the derivative of y with respect to x is given by:

dy/dx = e^(bx^2) * du/dx

Substituting the values of e^(bx^2) and du/dx, we have:

dy/dx = e^(bx^2) * 2bx

Therefore, the correct answer is dy/dx = 2bxe^(bx^2).

5) Calculate dy/dx when y = (x^2 + 2)e^(4x)

As given, we will differentiate the product of two functions u = x^2 + 2 and v = e^(4x).

Applying the product rule, the derivative of y with respect to x is given by:

dy/dx = (du/dx) * v + u * (dv/dx)

Taking the derivatives of u and v with respect to x, we have:

du/dx = 2x and dv/dx = 4e^(4x)

Substituting the values of du/dx, v, and dv/dx, we have:

dy/dx = (2x) * e^(4x) + (x^2 + 2) * (4e^(4x))

Simplifying, dy/dx = 2xe^(4x) + 4(x^2 + 2)e^(4x)

Therefore, the correct answer is dy/dx = 2xe^(4x) + 4(x^2 + 2)e^(4x).

1. To find dy/dx when y = ln(sinh(2x)), we can use the chain rule. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).

In this case, let f(u) = ln(u) and g(x) = sinh(2x).

First, find the derivative of f(u) with respect to u:
f'(u) = 1/u

Next, find the derivative of g(x) with respect to x:
g'(x) = d(sinh(2x))/dx

To find the derivative of sinh(2x), we can use the chain rule again. Let h(x) = 2x and f(t) = sinh(t).
Then, h'(x) = 2 and f'(t) = cosh(t).

Using the chain rule, the derivative of sinh(2x) with respect to x is:
d(sinh(2x))/dx = 2 * cosh(2x).

Now, multiply f'(g(x)) and g'(x) to get the derivative of y:
dy/dx = f'(g(x)) * g'(x) = (1/sinh(2x)) * (2 * cosh(2x)) = 2 * coth(2x).

Therefore, the correct answer is 2 * coth(2x).

2. To find dy/dx when sinh(3y) = cos(2x), let's solve this equation for y first.
sinh(3y) = cos(2x)
Take the inverse hyperbolic sine on both sides to isolate y:
3y = arcsinh(cos(2x))
Divide both sides by 3:
y = (1/3) * arcsinh(cos(2x))

Now, to find dy/dx, we can differentiate y with respect to x using the chain rule.

Let f(u) = (1/3) * arcsinh(u) and g(x) = cos(2x).

First, find the derivative of f(u) with respect to u:
f'(u) = (1/3) * (1/sqrt(1+u^2))

Next, find the derivative of g(x) with respect to x:
g'(x) = d(cos(2x))/dx

To find the derivative of cos(2x), we can use the chain rule again. Let h(x) = 2x and f(t) = cos(t).
Then, h'(x) = 2 and f'(t) = -sin(t).

Using the chain rule, the derivative of cos(2x) with respect to x is:
d(cos(2x))/dx = 2 * (-sin(2x)) = -2sin(2x).

Now, multiply f'(g(x)) and g'(x) to get the derivative of y:
dy/dx = f'(g(x)) * g'(x) = (1/3) * (1/sqrt(1+cos^2(2x))) * (-2sin(2x)) = -2sin(2x) / (3sqrt(1+cos^2(2x))).

Therefore, the correct answer is B. -2sin(2x) / (3sqrt(1+cos^2(2x))).

3. To find the derivative of y = sin^2(4x)cos(3x), we will use the product rule.

The product rule states that if y = f(x)g(x), then the derivative of y with respect to x (dy/dx) is given by:
dy/dx = f'(x)g(x) + f(x)g'(x).

In this case, let f(x) = sin^2(4x) and g(x) = cos(3x).

First, find the derivative of f(x) with respect to x:
f'(x) = d(sin^2(4x))/dx

To find the derivative of sin^2(4x), we can use the chain rule. Let h(x) = 4x and g(t) = sin^2(t).
Then, h'(x) = 4 and g'(t) = 2sin(t)cos(t).

Using the chain rule, the derivative of sin^2(4x) with respect to x is:
d(sin^2(4x))/dx = 4 * 2sin(4x)cos(4x) = 8sin(4x)cos(4x).

Next, find the derivative of g(x) with respect to x:
g'(x) = d(cos(3x))/dx

To find the derivative of cos(3x), we can use the chain rule. Let h(x) = 3x and g(t) = cos(t).
Then, h'(x) = 3 and g'(t) = -sin(t).

Using the chain rule, the derivative of cos(3x) with respect to x is:
d(cos(3x))/dx = 3 * (-sin(3x)) = -3sin(3x).

Now, multiply f'(x)g(x) and f(x)g'(x) and simplify to get the derivative of y:
dy/dx = f'(x)g(x) + f(x)g'(x) = 8sin(4x)cos(4x)cos(3x) - 3sin^2(4x)sin(3x).

Therefore, the correct answer is C. 8sin(4x)cos(4x)cos(3x) - 3sin^2(4x)sin(3x).

4. To find the derivative of y = e^(bx^2), we can use the chain rule.

The chain rule states that if y = f(g(x)), then the derivative of y with respect to x (dy/dx) is given by:
dy/dx = f'(g(x)) * g'(x).

In this case, let f(u) = e^u and g(x) = bx^2.

First, find the derivative of f(u) with respect to u:
f'(u) = e^u.

Next, find the derivative of g(x) with respect to x:
g'(x) = d(bx^2)/dx = 2bx.

Now, multiply f'(g(x)) and g'(x) to get the derivative of y:
dy/dx = f'(g(x)) * g'(x) = e^(bx^2) * 2bx = 2bxe^(bx^2).

Therefore, the correct answer is B. 2bxe^(bx^2).

5. To calculate dy/dx when y = (x^2 + 2)e^(4x), we can use the product rule.

The product rule states that if y = f(x)g(x), then the derivative of y with respect to x (dy/dx) is given by:
dy/dx = f'(x)g(x) + f(x)g'(x).

In this case, let f(x) = x^2 + 2 and g(x) = e^(4x).

First, find the derivative of f(x) with respect to x:
f'(x) = d/dx (x^2 + 2) = 2x.

Next, find the derivative of g(x) with respect to x:
g'(x) = d/dx (e^(4x)).

The derivative of e^(4x) can be found using the chain rule. Let h(x) = 4x and g(t) = e^t.
Then, h'(x) = 4 and g'(t) = e^t.

Using the chain rule, the derivative of e^(4x) with respect to x is:
d/dx (e^(4x)) = 4e^(4x).

Now, multiply f'(x)g(x) and f(x)g'(x) and simplify to get the derivative of y:
dy/dx = f'(x)g(x) + f(x)g'(x) = (2x) * e^(4x) + (x^2 + 2) * (4e^(4x)).

Therefore, the correct answer is D. dv/dx = (4xe^(4x)) + (x^2 + 2) * (4e^(4x)).