If y=xln(x)^2 , then dy/dx=
To find dy/dx, we can use the chain rule, which states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).
Let's apply the chain rule to the given function y = xln(x)^2.
First, we define f(u) = u^2 and g(x) = ln(x). Then, our function y can be rewritten as y = f(g(x)) = (ln(x))^2.
Next, we calculate the derivatives of f(u) and g(x):
f'(u) = 2u
g'(x) = 1/x [since the derivative of ln(x) with respect to x is 1/x]
Now, we can apply the chain rule:
dy/dx = f'(g(x)) * g'(x) = 2(ln(x)) * (1/x) = 2ln(x)/x.
Therefore, dy/dx = 2ln(x)/x.
To find the derivative of y = xln(x)^2 with respect to x, we can use the chain rule, which states that if we have a function y = f(g(x)), then its derivative dy/dx is given by dy/dx = f'(g(x)) * g'(x).
Let's break down the given expression to apply the chain rule: y = xln(x)^2
We can rewrite ln(x)^2 as (ln(x))^2. Now, let's consider g(x) = ln(x) and f(u) = u^2, where u = ln(x).
The first step is to find the derivative of f(u). The derivative of u^2 with respect to u is 2u.
Next, we find the derivative of g(x). The derivative of ln(x) with respect to x is 1/x.
Now, we can apply the chain rule: dy/dx = f'(g(x)) * g'(x)
dy/dx = 2u * (1/x)
Since u = ln(x), we substitute it back in: dy/dx = 2ln(x) * (1/x)
Simplifying further, dy/dx = 2ln(x)/x
So, the derivative of y = xln(x)^2 with respect to x is dy/dx = 2ln(x)/x.
dy/dx = x(2)(lnx)(1/x) + (lnx)^2
= 2lnx + (lnx)^2