Does anyone know how to find the derivative of 3x^ln(2x)?
d/dx f(x)g(x) =
f(x)g(x)* [g'(x)ln(f(x))+g(x)f'(x)/f(x) ]
3x^ln(2x) =
3 exp[ln(2x) ln(x)]
The derivative of exp(f(x)) is
exp(f(x)) f'(x)
You can easily prove this using the chain rule.
To find the derivative of the function 3x^ln(2x), you can use logarithmic differentiation. Logarithmic differentiation is a technique that allows you to find the derivative of a function that is not in a standard form like f(x) = a^x or f(x) = x^n.
Here's a step-by-step guide on how to find the derivative of 3x^ln(2x) using logarithmic differentiation:
Step 1: Take the natural logarithm (ln) of both sides of the equation to simplify the expression with the exponent:
ln(y) = ln(3x^ln(2x))
Step 2: Use the logarithmic properties to simplify the expression:
ln(y) = ln(3) + ln(x^ln(2x))
Step 3: Apply the power rule for logarithms, which states that ln(a^b) = b * ln(a):
ln(y) = ln(3) + (ln(2x) * ln(x))
Step 4: Differentiate both sides of the equation with respect to x using implicit differentiation. Remember that the derivative of ln(y) with respect to x is (1/y) * dy/dx, and the derivative of ln(x) with respect to x is 1/x:
(1/y) * dy/dx = 0 + (1 * ln(2x) * (1/x)) + (ln(2x) * 1/x * ln(x))
Step 5: Simplify the expression:
(1/y) * dy/dx = ln(2x)/x + ln(2x) * ln(x)/x
Step 6: To isolate dy/dx, multiply both sides of the equation by y:
dy/dx = y * (ln(2x)/x + ln(2x) * ln(x)/x)
Step 7: Replace y with the original expression:
dy/dx = 3x^ln(2x) * (ln(2x)/x + ln(2x) * ln(x)/x)
And there you have it! The derivative of 3x^ln(2x) is 3x^ln(2x) * (ln(2x)/x + ln(2x) * ln(x)/x).