To differentiate the function log₁(x) x², we first need to understand how logarithms and derivatives work.
The function log₁(x) can actually be simplified, as log base 1 is technically undefined for all positive numbers except at the value of 1. However, if we interpret the query as differentiating log(x) · x² instead, we can proceed with the differentiation.
The function can be rewritten as:
f(x) = log(x) · x²
To differentiate this product, we apply the product rule. The product rule states that if you have two functions u and v, then:
(u · v)’ = u’ · v + u · v’
Here, let:
u = log(x) and v = x²
First, we find the derivatives:
- u’ = 1/x (the derivative of log(x))
- v’ = 2x (the derivative of x²)
Now we can apply the product rule:
f'(x) = (1/x) · x² + log(x) · 2x
This simplifies to:
f'(x) = x + 2x log(x)
So, the derivative of log₁(x) x², interpreted as log(x) · x², is:
f'(x) = x + 2x log(x)