Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx(x log(x))
The derivative of y is y'(x):
y'(x) = d/dx(x log(x))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = log(x):
y'(x) = log(x) d/dx(x)+x d/dx(log(x))
The derivative of x is 1:
y'(x) = x (d/dx(log(x)))+1 log(x)
The derivative of log(x) is 1/x:
y'(x) = log(x)+1/x x
Simplify the expression:
y'(x) = 1+log(x)
Expand the left hand side:
Answer: | y'(x) = 1+log(x)
