Find the second derivative of y=loge(logex)
Show that y=loge(logex) is a solution of the equation
xd^2y/dx^2 + x(dy/dx)^2 + dy/dx=o
The first derivative of ln (ln x) is dy/dx = 1/(x*ln x)
The second derivative is:
d^2y/dx^2 = [-ln x/x^2 - (1/x^2)]/(ln x)^2
= -1/(x^2 ln x) -1/(x ln x)^2
x d^2y/dx^2 = -1/(x ln x) - (1/x)/lnx^2
It seems to fit the differential equation
To find the second derivative of y = loge(logex), we will apply the chain rule twice.
First, let's find the first derivative of y:
dy/dx = d/dx (loge(logex))
Using the chain rule, we can rewrite this as:
dy/dx = d/d(logex) (loge(u)) * d(logex)/dx
Here, u = logex. Differentiating with respect to u and then multiplying it by d(logex)/dx:
dy/dx = 1/u * (1/ex) * ex
Simplifying, we get:
dy/dx = 1/u
Now, let's find the second derivative of y:
d^2y/dx^2 = d/dx (dy/dx)
Using the chain rule again, we can rewrite this as:
d^2y/dx^2 = d/dx (1/u)
Applying the chain rule, we have:
d^2y/dx^2 = d(1/u)/du * du/dx
Here, du/dx = 1/ex. Differentiating 1/u with respect to u and multiplying by du/dx:
d^2y/dx^2 = -1/u^2 * (1/ex) * ex
Simplifying further:
d^2y/dx^2 = -1/(u^2 * ex)
Substituting u = logex:
d^2y/dx^2 = -1/(logex^2 * ex)
Now, let's show that y = loge(logex) is a solution to the given equation:
xd^2y/dx^2 + x(dy/dx)^2 + dy/dx = 0
Substituting the expressions we found earlier:
x * (-1/(logex^2 * ex)) + x * (1/u)^2 + 1/u = 0
Multiplying through by u^2 * ex to eliminate the fractions:
-x * ex + u^2 * ex + u = 0
Substituting u = logex:
-x * ex + (logex)^2 * ex + logex = 0
Rearranging terms:
(logex)^2 * ex - x * ex + logex = 0
This equation holds true since y = loge(logex) satisfies it. Therefore, y = loge(logex) is a solution to the equation:
xd^2y/dx^2 + x(dy/dx)^2 + dy/dx = 0.