Find the dy/dx

A. y = u sqrt u + 1;
u = 2x^2 - 2/3

B. x = u/ (1 + u^3);
y = u^2 /(1 + u^3)

Can someone help me

dy/dx= dy/du*du/dx

A y=u^3/2 +1
dy/du=3/2 sqrtu
du/dx=4x

dy/dx=3/2 sqrt(u)*4x then
dy/dx=3/2 sqrt(2x^2 -2/3)*4x

now on the second,
dy/dx=dy/du*du/dx=(dy/du)/(dx/du)

That is two messy to try to type in detail. Find dy/du,dx/du and put it in the fraction. Algebra to simplify might be a challenge.

x = u/(1 + u^3)

y = u^2/(1 + u^3)

dy/dx = dy/du / dx/du

dy/du = u(2-u^3)/(1+u^3)^2
dx/du = (1-2u^3)/(1+u^3)^2

dy/dx = u(2-u^3)/(1-2u^3)

Of course! I'll help you find the derivatives of the given functions.

A. To find dy/dx for the given function y = u(sqrt(u) + 1), we need to use the chain rule.

1. First, we need to find du/dx, as u is a function of x.
u = 2x^2 - 2/3

To find du/dx, we take the derivative of u with respect to x:
du/dx = d/dx (2x^2 - 2/3)

Taking the derivative, we get:
du/dx = 4x

2. Now, we can use the chain rule to find dy/dx. The chain rule states that if y = f(u) and u = g(x), then dy/dx = f'(u) * g'(x).

Let f(u) = u(sqrt(u) + 1) and g(x) = 2x^2 - 2/3.

Hence, dy/dx = f'(u) * g'(x).
f'(u) = d/du (u(sqrt(u) + 1))
= sqrt(u) + 1 + u(1/2sqrt(u))(1)
= sqrt(u) + 1 + u/2sqrt(u)
= sqrt(u) + 1 + u/(2u^(1/2))
= sqrt(u) + 1 + 1/(2u^(1/2))

Now, we substitute the values of u and du/dx back into the formula for dy/dx:
dy/dx = (sqrt(u) + 1 + 1/(2u^(1/2))) * du/dx
= (sqrt(2x^2 - 2/3) + 1 + 1/(2(2x^2 - 2/3)^(1/2))) * 4x

B. To find dy/dx for the given functions x = u/(1 + u^3) and y = u^2/(1 + u^3), we can use implicit differentiation.

1. Start by differentiating both sides of the equation x = u/(1 + u^3) with respect to x:
d/dx (x) = d/dx (u/(1 + u^3))

For the left side, the derivative of x with respect to x is simply 1.

For the right side, we use the quotient rule to differentiate u/(1 + u^3):
d/dx (u/(1 + u^3)) = (1 + u^3) * du/dx - u(3u^2) / (1 + u^3)^2

2. Now, we can solve for du/dx in terms of x. Rearrange the equation to isolate du/dx:
1 = (1 + u^3) * du/dx - 3u^3 / (1 + u^3)^2

Move the term with du/dx to the left side:
du/dx - (1 + u^3) * du/dx = 3u^3 / (1 + u^3)^2 - 1

Factor out du/dx:
du/dx (1 - (1 + u^3)) = 3u^3 / (1 + u^3)^2 - 1

Simplify:
-u^3 * du/dx = 3u^3 / (1 + u^3)^2 - 1

Divide both sides by -u^3:
du/dx = [3u^3 / (1 + u^3)^2 - 1] / -u^3

3. Now, we can find dy/dx by using the derived expression for du/dx. Recall that y = u^2/(1 + u^3).

dy/dx = d/dx (u^2/(1 + u^3))
= (1 + u^3) * d/dx (u^2) - u^2 * d/dx (1 + u^3) / (1 + u^3)^2

Differentiating u^2 and (1 + u^3) with respect to x:
= (1 + u^3) * 2u * du/dx - u^2 * 3u^2 * du/dx / (1 + u^3)^2

Substitute the expression for du/dx:
= (1 + u^3) * 2u * [3u^3 / (1 + u^3)^2 - 1] - u^2 * 3u^2 * [3u^3 / (1 + u^3)^2 - 1] / (1 + u^3)^2

Simplify and factor out u:
= 6u^4 / (1 + u^3) - 2u^2 - 9u^8 / (1 + u^3)^2 + 3u^6 / (1 + u^3)

Substitute the value of u back into the equation:
= 6(2x^2 - 2/3)^2 / (1 + (2x^2 - 2/3)^3) - 2(2x^2 - 2/3) - 9(2x^2 - 2/3)^8 / (1 + (2x^2 - 2/3)^3)^2 + 3(2x^2 - 2/3)^6 / (1 + (2x^2 - 2/3)^3)

Thus, we have found dy/dx for both given functions.