use logarithmic differentiation to find dy/dx for: [(x^2)(e^2)(x)] / [3√(2x-5)]
I assume you mean
y = [x^2 e^(2x)]/ [3√(2x-5)]
ln y = 2lnx + 2x - 3/2 ln(2x-5)
1/y y' = 2/x + 2 - 3/(2x-5)
1/y y' = (4x^2-9x-10)/[x(2x-5)]
y' = [x^2 e^(2x)]/ [3√(2x-5)] * (4x^2-9x-10)/[x(2x-5)]
x e^(2x) (4x^2-7x-10)
----------------------------
3(2x-5)^(3/2)
oops. That should have been
ln y = 2lnx + 2x - ln3 - 1/2 ln(2x-5)
and also 4x^2-7x-10 in the numerator
how did you get 4x^2-7x-10?
a little algebra. You didn't get that? Show whatcha got.
To find the derivative using logarithmic differentiation, follow these steps:
Step 1: Take the natural logarithm of both sides of the equation.
Step 2: Use logarithmic properties to simplify the expression.
Step 3: Differentiate implicitly.
Step 4: Solve for dy/dx.
Let's apply these steps to find dy/dx for the given function:
Step 1: Take the natural logarithm of both sides:
ln(y) = ln[(x^2)(e^2)(x)] - ln[3√(2x-5)]
Step 2: Simplify the expression using logarithmic properties:
ln(y) = ln(x^2) + ln(e^2) + ln(x) - ln[3√(2x-5)]
ln(y) = 2ln(x) + 2ln(e) + ln(x) - ln[3(2x-5)^(1/3)]
Note: ln(e) = 1, so we can simplify this further:
ln(y) = 2ln(x) + 2 + ln(x) - ln[3(2x-5)^(1/3)]
Step 3: Differentiate implicitly with respect to x:
(1/y) * (dy/dx) = 2 * (1/x) + 0 + (1/x) - (-1) * [1/3(2x-5)^(2/3)] * (6 - 0)
(1/y) * (dy/dx) = 3/x + 2/x + (2/(3√(2x-5)^2))
Step 4: Solve for dy/dx:
Multiplying both sides by y, we get:
dy/dx = y * [(3/x) + (2/x) + (2/(3√(2x-5)^2))]
Substituting the value of y back in:
dy/dx = [(x^2)(e^2)(x)] / [3√(2x-5)] * [(3/x) + (2/x) + (2/(3√(2x-5)^2))]
Simplifying further, we get the final expression for dy/dx:
dy/dx = (x^2)(e^2)(x) * [(3 + 2)(√(2x-5)) + 2] / [3(x(2x-5)^(1/3))]