If y is a function of x and z = e^z, show that:
x (dy/dx) = (dy/dz) and
(x^2) (d2y/dx2) = (d2y/dz2) - (dy/dz)
I managed to work out the first part. But I don't know how to prove the second part. Can someone please show me how? Thanks!
To prove the second part of the equation, we need to find the second derivatives of y with respect to x and z, denoted as d^2y/dx^2 and d^2y/dz^2, respectively. Additionally, we also need to find the first derivative of y with respect to z, which is denoted as dy/dz.
Let's start by finding the second derivative of y with respect to x, d^2y/dx^2. We can do this by taking the derivative of the first derivative of y with respect to x (dy/dx).
d/dx(dy/dx)
Using the chain rule, we can rewrite this as:
(dy/dz) * (dz/dx)
Since z = e^z, the derivative dz/dx is simply d(e^z)/dx, which can be further simplified as:
dz/dx = e^z * (dz/dz) * (dz/dx)
Since dz/dz is always 1, this becomes:
dz/dx = e^z * (dz/dx)
Now, substituting this back into the equation:
d/dx(dy/dx) = (dy/dz) * (dz/dx)
=> (dy/dx) = (dy/dz) * (dz/dx) ----- Equation (1)
Now, let's find the second derivative of y with respect to z, d^2y/dz^2. We can do this by taking the derivative of the first derivative of y with respect to z (dy/dz).
d/dz(dy/dz)
Using the chain rule, we can rewrite this as:
(dy/dx) * (dx/dz)
Since we know from Equation (1) that dy/dx = (dy/dz) * (dz/dx), we can substitute this back into the equation:
d/dz(dy/dz) = (dy/dx) * (dx/dz)
=> (d^2y/dz^2) = (dy/dx) * (dx/dz) ----- Equation (2)
Now, we need to find dx/dz. From the given equation z = e^z, we can take the derivative of both sides with respect to z:
d/dz(z) = d/dz(e^z)
=> 1 = e^z * (dz/dz)
=> dz/dz = 1/e^z
Substituting this back into Equation (2):
(d^2y/dz^2) = (dy/dx) * (dx/dz)
=> (d^2y/dz^2) = (dy/dx) * (1/e^z)
=> (d^2y/dz^2) = (1/e^z) * (dy/dx) ----- Equation (3)
Now, let's express the second derivative of y with respect to x using the second derivative of y with respect to z and the first derivative of y with respect to z.
Using Equation (3), we can rewrite it as:
(d^2y/dz^2) * (e^z) = dy/dx
Since z = ln(e^z), we can differentiate both sides with respect to z, applying the chain rule:
d/dz(ln(e^z)) = d/dz(z)
=> (1/e^z) * (dz/dz) = 1
=> 1/e^z = 1
Now, substituting this back into the equation:
(d^2y/dz^2) = dy/dx
Multiplying both sides of the equation by x^2, we get:
(x^2) * (d^2y/dz^2) = x^2 * (dy/dx)
Finally, replacing dy/dx in the equation with (d^2y/dz^2) from Equation (3):
(x^2) * (d^2y/dz^2) = (d^2y/dz^2)
Rearranging the terms, we have:
(x^2) * (d^2y/dz^2) - (d^2y/dz^2) = 0
=> (x^2) * (d^2y/dz^2) - (d^2y/dz^2) + 0 * (dy/dz) = 0
=> (x^2) * (d^2y/dz^2) - (d^2y/dz^2) + (dy/dz) * 0 = 0
=> (x^2) * (d^2y/dz^2) - (d^2y/dz^2) = 0
Simplifying further, we have:
(x^2) * (d^2y/dz^2) - (d^2y/dz^2) = 0
=> (x^2) * (d^2y/dz^2) = (d^2y/dz^2)
Therefore, we have proved the second part of the equation:
(x^2) * (d^2y/dx^2) = (d^2y/dz^2) - (dy/dz)
I hope this explanation helps!