How do you use logarithmic differentiation to find the derivative of #y=(tanx)^(1/x)#?

Question

Emilia Aguilar · Accepted Answer

#y=(tanx)^(1/x)# #lny=ln((tanx)^(1/x))# #lny=1/xln(tanx)#  Differentiate implicitly. #1/y (dy)/(dx)=-1/x^2ln(tan(x))+1/x(1/tan(x) sec^2(x))# At this point it's nice to simplify using #1/tanx=cotx# and #cotx * secx=cscx# So,
#1/y (dy)/(dx)=-1/x^2ln(tan(x))+1/x(cscx secx)# # (dy)/(dx)=y(-1/x^2ln(tan(x))+1/x(cscx secx))# # (dy)/(dx) =( tanx)^(1/x)(-1/x^2ln(tan(x))+1/x(cscx secx))#  #" "# (Not pretty, but correct.) # (dy)/(dx) =( tanx)^(1/x)(x(cscx secx)-ln(tan(x)))*1/x^2# #" "# (Isn't a whole lot better.)

Adam Adkins · Answer

undefined To use logarithmic differentiation to find the derivative of $ y = (\tan x)^{\frac{1}{x}} $, you would follow these steps:

1. Take the natural logarithm of both sides: $ \ln y = \ln \left( (\tan x)^{\frac{1}{x}} \right) $.

2. Apply the logarithm property $ \ln(a^b) = b \ln(a) $ to simplify: $ \ln y = \frac{1}{x} \cdot \ln(\tan x) $.

3. Differentiate both sides with respect to $ x $ implicitly: $ \frac{1}{y} \cdot \frac{dy}{dx} = -\frac{1}{x^2} \cdot \ln(\tan x) + \frac{1}{x} \cdot \sec^2 x $.

4. Solve for $ \frac{dy}{dx} $: $ \frac{dy}{dx} = y \left( -\frac{1}{x^2} \cdot \ln(\tan x) + \frac{1}{x} \cdot \sec^2 x \right) $.

5. Substitute back $ y = (\tan x)^{\frac{1}{x}} $: $ \frac{dy}{dx} = (\tan x)^{\frac{1}{x}} \left( -\frac{1}{x^2} \cdot \ln(\tan x) + \frac{1}{x} \cdot \sec^2 x \right) $.