Question

In Exercises \(31-42,\) find \(d y / d x\). $$y=3\left(2 x^{-1 / 2}+1\right)^{-1 / 3}$$

Short answer

The derivative of \(y\) with respect to \(x\) is \(dy/dx = x^{-3/2}(2x^{-1/2} + 1)^{-4/3}\)

Step-by-step solution

Identify the outer and inner functions

The outer function is \( f(u) = 3u^{-1/3} \), and the inner function \( u = 2x^{-1/2} + 1 \). The derivative will make use of chain rule which requires differentiating both the inner and outer functions.

Differentiate the outer function

Keep \( u = 2x^{-1/2} + 1 \) constant while differentiating. \n For \( f(u) = 3u^{-1/3} \), apply the rule for derivative of \(x^n\), giving \(f'(u) = 3(-1/3)u^{-1/3-1} = -u^{-4/3}\).

Differentiate the inner function

Keep the outer function constant while differentiating: \n For \(u = 2x^{-1/2} + 1 \), the derivative is \(u'(x) = 2(-1/2)x^{-1/2-1} = -x^{-3/2}\).

Multiply the derivatives using the chain rule

According to chain rule, \(dy/dx = f'(u) \cdot u'(x)\). Substitute the differentiated outer and inner functions from steps 2 and 3 to obtain: \(dy/dx = -u^{-4/3} \cdot -x^{-3/2}\).

Substitute the inner function

Substitute \( u = 2x^{-1/2} + 1 \) back into the derivative to obtain the final form of \(dy/dx\).