If y = (ln(x-2))/(3e^x, then dy/dx =
To find the derivative of y with respect to x, we can use the quotient rule. The quotient rule states that if we have a function in the form of f(x)/g(x), where f(x) and g(x) are differentiable functions, then the derivative is given by (g(x)f'(x) - f(x)g'(x))/(g(x))^2.
In this case, let's define f(x) = ln(x-2) and g(x) = 3e^x. Now we can differentiate f(x) and g(x) separately.
First, let's differentiate f(x), which is ln(x-2). The derivative of ln(u) is 1/u multiplied by the derivative of u. Applying this rule, the derivative of ln(x-2) is 1/(x-2) multiplied by the derivative of (x-2), which is 1.
Next, let's differentiate g(x), which is 3e^x. The derivative of e^u is e^u multiplied by the derivative of u. Since u = x, the derivative of e^x is e^x.
Now we have the derivative of f(x) and g(x), which are f'(x) = 1/(x-2) and g'(x) = 3e^x, respectively.
Using the quotient rule, we can find dy/dx as:
(dy/dx) = (g(x)f'(x) - f(x)g'(x))/(g(x))^2
= (3e^x * 1/(x-2) - ln(x-2) * 3e^x) / (3e^x)^2
= (3e^x/(x-2) - 3e^x*ln(x-2)) / (9e^(2x))
Therefore, dy/dx = (3e^x/(x-2) - 3e^x*ln(x-2)) / (9e^(2x)).