Question

Find the derivative of the algebraic function. $$ f(x)=x\left(1-\frac{4}{x+3}\right) $$

Short answer

The derivative of the function is \( f'(x) = 1 - \frac{4}{x+3} + \frac{4x}{(x+3)^2}\).

Step-by-step solution

Simplify the Function

Start by simplifying the function \(f(x)\). This makes it easier to find the derivative. \[ f(x)=x\left(1-\frac{4}{x+3}\right)=x-\frac{4x}{x+3}.\]

Apply the Product and Quotient Rules

Derive the two parts of the function separately. The derivative of \(x\) is \(1\). For the second part, let \(u = x\) and \(v = \frac{4}{x+3}\). Using the product rule, the derivative of this part is \[ u'v + uv' = 1\cdot\frac{4}{x+3} + x\cdot\left(\frac{0(x+3) - 4(1)}{(x+3)^2}\right) = \frac{4}{x+3} - \frac{4x}{(x+3)^2}.\]

Simplify the Derivative

Combine the two parts to get the final derivative: \[ f'(x) = 1 - \frac{4}{x+3} + \frac{4x}{(x+3)^2}.\]