Compute the derivatives f(x)=ln ((2x)/(x+1))^(3)

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Compute the derivatives   f(x)=ln ((2x)/(x+1))^(3)

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Apply Chain Rule for ln(u): Apply the chain rule to the logarithmic function. The derivative of ln(u) is 1/u * du/dx, where u is a function of x. In this case, u = ((2x)/(x+1))^3.Differentiate Inside Function u: Differentiate the inside function u = ((2x)/(x+1))^3. To differentiate u, we will use the chain rule again, since u is a composition of two functions: the cube function and the division of 2x by x+1.Differentiate Cube Function: Differentiate the cube function. The derivative of v^3 with respect to v is 3v^2. Let v = (2x)/(x+1), then the derivative of v^3 with respect to x is 3v^2 * dv/dx.Differentiate v = (2x)/(x+1): Differentiate v = (2x)/(x+1). To differentiate v, we will use the quotient rule: (v'/u - vu'/u^2), where v = 2x and u = x+1. The derivative of 2x is 2, and the derivative of x+1 is 1.Apply Quotient Rule for dv/dx: Apply the quotient rule to find dv/dx. dv/dx = (2*(x+1) - 2x*1)/(x+1)^2 = (2x + 2 - 2x)/(x+1)^2 = 2/(x+1)^2.Combine Results from Step 3 and Step 5: Combine the results from Step 3 and Step 5. The derivative of u with respect to x is 3v^2 * dv/dx = 3*((2x)/(x+1))^2 * 2/(x+1)^2.Simplify Expression for du/dx: Simplify the expression for du/dx. 3*((2x)/(x+1))^2 * 2/(x+1)^2 = 3*(4x^2/(x+1)^2) * 2/(x+1)^2 = 24x^2/(x+1)^4.Apply Chain Rule for f(x): Apply the chain rule to find the derivative of f(x). The derivative of f(x) with respect to x is 1/u * du/dx = 1/((2x)/(x+1))^3 * 24x^2/(x+1)^4.Simplify Expression for f'(x): Simplify the expression for f'(x). 1/((2x)/(x+1))^3 * 24x^2/(x+1)^4 = 24x^2/((2x)^3) * 1/(x+1) = 24x^2/(8x^3) * 1/(x+1).Further Simplify f'(x): Further simplify the expression for f'(x). 24x^2/(8x^3) * 1/(x+1) = 3/x * 1/(x+1) = 3/(x(x+1)).

Compute the derivatives $\newline$ $f(x)=\ln \left(\frac{2x}{x+1}\right)^3$

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Compute the derivatives $\newline$ $f(x)=\ln \left(\frac{2x}{x+1}\right)^3$