Let #y=[(x^(2x)(x-1)^3)/(3+5x)^4]#, how do you use logarithmic differentiation to find #dy/dx#?

Question

Emily Ammerman · Accepted Answer

# dy/dx= (x^(2x)(x-1)^3)/(3+5x)^4{2 + 2lnx + 3/(x-1)-20/(3+5x)} #

#y=[(x^(2x)(x-1)^3)/(3+5x)^4]#

Taking (natural) logs of both sides:

# ln y=ln[(x^(2x)(x-1)^3)/(3+5x)^4] # # \ \ \ \ \ \=ln \ x^(2x) + ln\ (x-1)^3 - ln \ (3+5x)^4 # # \ \ \ \ \ \=2xln \ x + 3ln\ (x-1) - 4ln \ (3+5x) #

Differentiating the LHS implicitly and the RHS using the product rule gives: # 1/y dy/dx \ = (2x)(1/x) + (2)(lnx) + 3/(x-1)-4*5/(3+5x) # # \ \ \ \ \ \ \ \ \ \ = 2 + 2lnx + 3/(x-1)-20/(3+5x) # # :. dy/dx= (x^(2x)(x-1)^3)/(3+5x)^4{2 + 2lnx + 3/(x-1)-20/(3+5x)} #

Luke Alvarez · Answer

undefined To find $ \frac{dy}{dx} $ using logarithmic differentiation, follow these steps:

1. Take the natural logarithm of both sides of the equation: 
$$ \ln(y) = \ln\left(\frac{x^{2x}(x-1)^3}{(3+5x)^4}\right) $$

2. Apply logarithmic properties to simplify the expression:
$$ \ln(y) = \ln(x^{2x}) + \ln((x-1)^3) - \ln((3+5x)^4) $$

3. Use the properties of logarithms to bring down the exponents as coefficients:
$$ \ln(y) = (2x)\ln(x) + 3\ln(x-1) - 4\ln(3+5x) $$

4. Differentiate both sides of the equation with respect to $ x $:
$$ \frac{1}{y} \frac{dy}{dx} = 2\ln(x) + 2x \cdot \frac{1}{x} + 3 \cdot \frac{1}{x-1} - 4 \cdot \frac{1}{3+5x} \cdot 5 $$

5. Solve for $ \frac{dy}{dx} $:
$$ \frac{dy}{dx} = y \left( 2\ln(x) + 2 + \frac{3}{x-1} - \frac{20}{3+5x} \right) $$

6. Substitute the expression for $ y $ back into the equation:
$$ \frac{dy}{dx} = \left( \frac{x^{2x}(x-1)^3}{(3+5x)^4} \right) \left( 2\ln(x) + 2 + \frac{3}{x-1} - \frac{20}{3+5x} \right) $$

That's the derivative $ \frac{dy}{dx} $ using logarithmic differentiation.