Find the derivatives of

1. ln[(2x^3⋅2e^x)]

2. xy = cot(xy)

3. y = cos(x^3 + 27)

I don't really know where to start with these.
The answer is (3+x)/(x) for the first, y/-x for the second, and 35. -3x^2 sin(x^3 + 27) but I have no idea how they got there.
Thank you!

a. y=ln(uv)

y'=1/(uv)*(uv'+vu')

ok, u=2x^3 u'=6x^2
so y'=1/(2x^3*2e^x) * (2x^3*2e^x + 6x^2*2e^x(
y'=(3+x)/x

Thanks so much. Could you help me with 2 & 3?

just do the same stuff:

xy = cot(xy)
y + xy' = -csc^2(xy) (y + xy')
y'(x + xcsc^2(xy)) = -ycsc^2(xy)-y
xy' (1+csc^2(xy)) = -y(1+csc^2(xy))
xy' = -y
y' = -y/x

The other is just a simple chain rule.
y = cos(u), so
y' = -sin(u) u'

Sure! I'd be happy to explain how to find the derivatives of these functions.

1. To find the derivative of ln[(2x^3⋅2e^x)], we can use the chain rule. The chain rule states that for a function f(g(x)), the derivative is given by f'(g(x)) * g'(x).

In our case, f(u) = ln(u) and g(x) = (2x^3⋅2e^x). We can think of ln[(2x^3⋅2e^x)] as f(g(x)). So, to find the derivative, we need to find f'(g(x)) and g'(x) separately.

First, let's find f'(u). The derivative of ln(u) is 1/u. So, f'(u) = 1/u.

Now, let's find g'(x). We can break down g(x) as follows:
g(x) = (2x^3⋅2e^x) = 2x^3 * 2e^x

To find the derivative of 2x^3, we can apply the power rule. The power rule states that the derivative of x^n, where n is a constant, is nx^(n-1). So, the derivative of 2x^3 is 6x^2.

To find the derivative of 2e^x, we can use the exponential rule. The exponential rule states that the derivative of e^x is e^x. Since we have a constant multiple (2), the derivative of 2e^x is 2e^x.

Now, we have f'(g(x)) = 1/(2x^3 * 2e^x) and g'(x) = 6x^2 * 2e^x.

Applying the chain rule, the derivative of ln[(2x^3⋅2e^x)] is (1/(2x^3 * 2e^x)) * (6x^2 * 2e^x). Simplifying this expression gives us (3+x)/x.

2. To find the derivative of xy = cot(xy), we can use implicit differentiation. Implicit differentiation is used when the given equation cannot be easily solved for y.

Start by differentiating both sides of the equation with respect to x. The derivative of xy with respect to x is y + x(dy/dx), and the derivative of cot(xy) with respect to x can be found using the chain rule.

To differentiate cot(xy), we need to differentiate cot(u), where u = xy. The derivative of cot(u) is -csc^2(u), and by the chain rule, we multiply this by the derivative of u with respect to x, which is y + x(dy/dx).

So, we have y + x(dy/dx) = -csc^2(xy) * (y + x(dy/dx)).

Now, we can solve for dy/dx by isolating it:
y + x(dy/dx) = -csc^2(xy) * (y + x(dy/dx))
y + x(dy/dx) = -y*csc^2(xy) - x(dy/dx)*csc^2(xy)
x(dy/dx) + x(dy/dx)*csc^2(xy) = -y*csc^2(xy) - y

Factor out dy/dx:
x(dy/dx)(1 + csc^2(xy)) = -y*csc^2(xy) - y

Divide both sides by (x(1 + csc^2(xy))):
dy/dx = (-y*csc^2(xy) - y) / (x(1 + csc^2(xy)))

Simplifying this expression further, we get dy/dx = y/(-x).

3. To find the derivative of y = cos(x^3 + 27), we can use the chain rule.

The derivative of cos(u) with respect to u is -sin(u). Let's denote u as x^3 + 27. So, the derivative of cos(x^3 + 27) with respect to x is -sin(x^3 + 27).

To apply the chain rule, we multiply this by the derivative of x^3 + 27 with respect to x. The derivative of x^3 with respect to x is 3x^2, and the derivative of 27 (which is a constant) with respect to x is 0. So, the derivative of x^3 + 27 with respect to x is 3x^2.

Multiplying these two derivatives, we get (-sin(x^3 + 27)) * (3x^2).

Therefore, the derivative of y = cos(x^3 + 27) is -3x^2*sin(x^3 + 27).