f(x)=3x^ln(x),
find f '(9).
i can't find the answer :(
I got 3x^lnx(((ln(lnx))/x)+ 1/lnx)
F(x) = 3X^ln(x).
F'(x) = (3lnx)(X^(lnx-1),
F'(9) = (3*ln9)(9^(ln9-1),
F'(9) = (3*2.1972)(9^(2.1972-1),
F'(9) = (6.5917)(9^1.1972),
F'(9) = (6.5917)(13.8817),
F'(9) = 91.5040.
To find the derivative of f(x) = 3x^ln(x), you can use the product rule of differentiation. The product rule states that if you have two functions u(x) and v(x), the derivative of their product uv with respect to x is given by:
(d/dx) (u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
In this case, u(x) = 3x and v(x) = x^ln(x). Now, let's differentiate each part separately.
First, let's find the derivative of u(x) = 3x:
u'(x) = 3
Now, let's find the derivative of v(x) = x^ln(x). To do this, we need to use logarithmic differentiation. We take the natural logarithm of both sides of v(x):
ln(v(x)) = ln(x^ln(x))
Using the properties of logarithms, we can simplify this expression:
ln(v(x)) = ln(x) * ln(x)
Now, let's differentiate both sides of the equation with respect to x:
(d/dx) ln(v(x)) = (d/dx) (ln(x) * ln(x))
On the left side, we can use the chain rule:
(1/v(x)) * v'(x) = (ln(x) * (1/x) + (1/x) * ln(x))
Since v(x) = x^ln(x), we can substitute it back in:
(1/x^ln(x)) * v'(x) = ln(x)/x + ln(x)/x
Now, let's solve for v'(x):
v'(x) = (x^ln(x)) * [ln(x)/x + ln(x)/x]
Finally, we can use the product rule:
f'(x) = u'(x) * v(x) + u(x) * v'(x)
= 3 * x^ln(x) + 3x * [ln(x)/x + ln(x)/x]
Now, to find f'(9), we substitute x = 9 into the expression for f'(x):
f'(9) = 3 * 9^ln(9) + 3 * 9 * [ln(9)/9 + ln(9)/9]
Unfortunately, calculating this expression precisely would require knowing the exact value of ln(9), which is not a whole number. So, without further information, we cannot determine the exact numerical value of f'(9).