Question

Differentiate (x-2/x+2)^3/4

Answer 1

To differentiate the function \(\left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}\), we will use the chain rule. Let:

\[
y = \left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}
\]

First, we can differentiate \(y\) with respect to \(u\), where:

\[
u = \frac{x-2}{x+2}
\]

Thus, we have:

\[
y = u^{\frac{3}{4}}
\]

Using the chain rule, we differentiate \(y\):

\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
\]

1. **Differentiate \(y\) with respect to \(u\)**:

\[
\frac{dy}{du} = \frac{3}{4}u^{-\frac{1}{4}}
\]

2. **Differentiate \(u\) with respect to \(x\)** using the quotient rule, where \(f(x) = x - 2\) and \(g(x) = x + 2\):

\[
\frac{du}{dx} = \frac{g(x) \frac{df}{dx} - f(x) \frac{dg}{dx}}{(g(x))^2}
\]

Calculating:

- \(f'(x) = 1\)
- \(g'(x) = 1\)

So:

\[
\frac{du}{dx} = \frac{(x+2)(1) - (x-2)(1)}{(x+2)^2} = \frac{x + 2 - (x - 2)}{(x + 2)^2}
\]

This simplifies to:

\[
\frac{du}{dx} = \frac{4}{(x + 2)^2}
\]

3. **Combine the derivatives**:

Now we can substitute \(\frac{dy}{du}\) and \(\frac{du}{dx}\):

\[
\frac{dy}{dx} = \frac{3}{4}u^{-\frac{1}{4}} \cdot \frac{4}{(x + 2)^2}
\]

Recall that \(u = \frac{x - 2}{x + 2}\), so \(u^{-\frac{1}{4}} = \left(\frac{x-2}{x+2}\right)^{-\frac{1}{4}}\).

Thus, we have:

\[
\frac{dy}{dx} = \frac{3}{4} \cdot \left(\frac{x-2}{x+2}\right)^{-\frac{1}{4}} \cdot \frac{4}{(x + 2)^2}
\]

The \(4\) in the numerator and denominator cancels out:

\[
\frac{dy}{dx} = 3 \cdot \left(\frac{x-2}{x+2}\right)^{-\frac{1}{4}} \cdot \frac{1}{(x + 2)^2}
\]

Thus, the final expression for the derivative is:

\[
\frac{dy}{dx} = \frac{3}{(x + 2)^2} \left(\frac{x - 2}{x + 2}\right)^{-\frac{1}{4}}
\]

or equivalently:

\[
\frac{dy}{dx} = \frac{3}{(x + 2)^2} \left(\frac{x + 2}{x - 2}\right)^{\frac{1}{4}}
\]

This is the derivative of \(\left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}\).