If f(x,y)=0, phi(y,z)=0, show that partial differential of f wrt y •partial differential of phi wrt z • partial differential of z wrt x =partial differential of f wrt x • partial differential of phi wrt y
To show that the equation
∂f/∂y * ∂ϕ/∂z * ∂z/∂x = ∂f/∂x * ∂ϕ/∂y,
we'll use the chain rule for partial derivatives. Let's start by taking the partial derivative of the function f(x, y) with respect to y:
∂f/∂y.
Next, let's take the partial derivative of the function ϕ(y, z) with respect to z:
∂ϕ/∂z.
Now, let's take the partial derivative of z with respect to x:
∂z/∂x.
By the chain rule, the derivative of f with respect to x can be expressed as:
∂f/∂x = (∂f/∂y) * (∂y/∂x).
Similarly, the derivative of ϕ with respect to y can be expressed as:
∂ϕ/∂y = (∂ϕ/∂z) * (∂z/∂y).
Now, substituting these expressions in the equation to be proved, we have:
(∂f/∂y) * (∂ϕ/∂z) * (∂z/∂x) = (∂f/∂x) * (∂ϕ/∂y).
Using the chain rule expressions, this equation can be rewritten as:
(∂f/∂y) * (∂ϕ/∂z) * (∂z/∂x) = (∂f/∂y) * (∂y/∂x) * (∂ϕ/∂z) * (∂z/∂y).
Now, since the equation f(x, y) = 0 holds, the left-hand side of the equation becomes 0. Therefore, (∂f/∂y) = 0.
Substituting this result back into the equation, we have:
0 = 0 * (∂y/∂x) * (∂ϕ/∂z) * (∂z/∂y).
Simplifying further, we get:
0 = (∂f/∂x) * (∂ϕ/∂y).
Thus, we have shown that:
∂f/∂y * ∂ϕ/∂z * ∂z/∂x = ∂f/∂x * ∂ϕ/∂y.
Therefore, the desired equation is proved.
To show the given equation, we need to use the chain rule in differential calculus. The chain rule states that if we have a composition of functions, we can find the derivative of the entire composition by multiplying the derivatives of the individual functions.
Let's differentiate both sides of the equation f(x, y) = 0 with respect to x:
∂f/∂x + ∂f/∂y * ∂y/∂x = 0 -- (1)
Next, differentiate the equation φ(y, z) = 0 with respect to y:
∂φ/∂y + ∂φ/∂z * ∂z/∂y = 0 -- (2)
Now, let's multiply both sides of equation (1) by ∂φ/∂z and both sides of equation (2) by ∂f/∂y:
(∂f/∂x) * (∂φ/∂z) + (∂f/∂y) * (∂y/∂x) * (∂φ/∂z) = 0 -- (3)
(∂φ/∂y) * (∂f/∂y) + (∂φ/∂z) * (∂z/∂y) * (∂f/∂y) = 0 -- (4)
Notice that (∂y/∂x) and (∂z/∂y) are just the partial derivatives of y with respect to x and z respectively. We can now replace them in equations (3) and (4) accordingly:
(∂f/∂x) * (∂φ/∂z) + (∂f/∂y) * (∂y/∂x) * (∂φ/∂z) = 0 -- (5)
(∂φ/∂y) * (∂f/∂y) + (∂φ/∂z) * (∂z/∂y) * (∂f/∂y) = 0 -- (6)
Comparing equations (5) and (6), we can observe that they are the same. Thus,
(∂f/∂x) * (∂φ/∂z) = (∂φ/∂y) * (∂f/∂y) -- (7)
This is the desired result, which shows that the partial derivatives of f and φ satisfy the given equation:
∂f/∂x * ∂φ/∂z * ∂z/∂x = ∂f/∂x * ∂φ/∂y.