Question

In Exercises \(1-28\) , find \(d y / d x\) . Remember that you can use NDER to support your computations. $$y=1 / \log _{2} x$$

Short answer

The derivative of \(y = \frac{1}{\log_2{x}}\) is \( \frac{d y}{d x} = -\frac{1}{x\log_{2} x} \)

Step-by-step solution

Rewrite the Expression with Base \(e\)

First, change the logarithm with base 2 to a natural logarithm as it will simplify the process of finding the derivative. Use the change-of-base formula: \(log_b{a} = \frac{\ln a}{\ln b} \). So, the equation becomes \(y = \frac{\ln x}{\ln 2}\).

Differentiate the Function

Differentiate the function using the quotient rule for differentiation, which stipulates for the function \(u/v\) that \((u/v)' = (v * u' - u * v') / v²\). Here, \(u = \ln{x}\) whose derivative \(u' = 1/x\), and \(v = \ln{2}\) whose derivative \(v' = 0\) as \(\ln{2}\) is a constant. Substituting these in the formula: \( \frac{d y}{d x} = \left(\frac{\ln{2} * \frac{1}{x} - \frac{\ln{x} * 0}{\ln{2}^2}\right) = \frac{1}{x\ln{2}}\).

Simplify the Expression

Now, distinguish that this equals to the reciprocal of \(\log_2 x\). Lastly, as the original expression was \(y = \frac{1}{\log_2 x}\), the final derivative of \(y\) would be the opposite of that, so, \( \frac{d y}{d x} = -\frac{1}{x\log_{2} x} \)