find f'(x)
-4x^2 lnx
just use the product rule:
f = -4x^2 * lnx
f' = -8x * lnx + (-4x^2)*(1/x)
= -8x*lnx - 4x
= -4x(2lnx + 1)
To find the derivative of the function f(x) = -4x^2 ln(x), we can use the product rule and the chain rule.
The product rule states that if we have a function u(x) multiplied by another function v(x), the derivative of the product is given by:
(uv)' = u'v + uv'
In this case, u(x) = -4x^2 and v(x) = ln(x).
To find u'(x), we apply the power rule, which states that the derivative of x^n, where n is any real number, is given by:
d/dx (x^n) = nx^(n-1)
So, u'(x) = d/dx (-4x^2) = -8x.
To find v'(x), we apply the chain rule, which states that if we have a function g(x) inside another function f(x), the derivative of f(g(x)) is given by:
d/dx (f(g(x))) = f'(g(x)) * g'(x)
In this case, f(x) = ln(x). The derivative of ln(x) is given by:
d/dx (ln(x)) = 1/x
Therefore, v'(x) = d/dx (ln(x)) = 1/x.
Now, using the product rule, we can find the derivative of f(x):
f'(x) = u'v + uv'
= (-8x)(ln(x)) + (-4x^2)(1/x)
= -8xln(x) - 4x
Hence, the derivative of f(x) = -4x^2 ln(x) is f'(x) = -8xln(x) - 4x.