Question

Group Activity In Exercises \(43-48,\) use the technique of logarithmic differentiation to find \(d y / d x\) . $$y=x^{\tan x}, x>0$$

Short answer

\(\frac{dy}{dx} = x^{\tan x}(\sec^2x \ln x + \frac{\tan x}{x})\)

Step-by-step solution

Apply Logarithm to both sides

Starting with \(y=x^{\tan x}\), we apply logarithm to both sides which gives us: \[\ln y = \tan x \ln x\]

Differentiate both sides

Differentiating both sides with respect to \(x\), using the product and chain rules on the right and applying the formula \(\frac{d}{dx} \ln u = \frac{u'}{u}\) on the left gives us:\[\frac{1}{y} \frac{dy}{dx} = \sec^2x \ln x + \frac{\tan x}{x}\]

Isolate dy/dx

Multiply both sides by \(y\), which isolates the derivative on the left hand side:\[\frac{dy}{dx} = y(\sec^2x \ln x + \frac{\tan x}{x})\]

Substitute y with the original function

Finally, replacing \(y\) with the original function \(y=x^{\tan x}\) in the expression for \(\frac{dy}{dx}\) yields:\[\frac{dy}{dx} = x^{\tan x}(\sec^2x \ln x + \frac{\tan x}{x})\]