Let f(x) = 3(x^6)lnx. f'(x)= ...
Use the product rule for derivatives:
d(uv) dx = v du/dx + u dv/dx
Let u(x) = 3x^6 and v(x) = ln x
Take it from there. I am sure you know what du/dx and dv/dx are.
To find the derivative of f(x) = 3(x^6)lnx, you can use the product rule and the chain rule.
Step 1: Apply the product rule
The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is given by:
(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)
In this case, let u(x) = 3(x^6) and v(x) = lnx. Then we have:
f'(x) = (d/dx)(3(x^6)lnx) = u'(x)v(x) + u(x)v'(x)
Step 2: Find the derivatives of u(x) and v(x)
To differentiate u(x) = 3(x^6), we can use the power rule, which states that if you have a function of the form f(x) = ax^n, then its derivative is given by:
f'(x) = nax^(n-1)
Applying the power rule to u(x):
u'(x) = d/dx(3(x^6)) = 6(3)x^5 = 18x^5
To differentiate v(x) = lnx, we can use the chain rule, which states that if you have a composite function, the derivative is the derivative of the outer function multiplied by the derivative of the inner function.
Applying the chain rule to v(x):
v'(x) = d/dx(lnx) = 1/x
Step 3: Substitute the derivatives into f'(x)
Now we can substitute the derivatives we found into the expression for f'(x):
f'(x) = u'(x)v(x) + u(x)v'(x) = 18x^5 * lnx + 3(x^6) * (1/x)
Simplifying this expression, we get:
f'(x) = 18x^5lnx + 3x^5 = 3x^5(6lnx + 1)
Therefore, the derivative of f(x) = 3(x^6)lnx is f'(x) = 3x^5(6lnx + 1).