Finding the Derivative of $ y = x^{\tan(x)} $ at $ x = \frac{\pi}{4} $

This solution explains how to find the derivative of the function $ y = x^{\tan(x)} $ and evaluate it at a specific point, $ x = \frac{\pi}{4} $. This type of function, where the base and exponent both contain the variable, is typically solved using logarithmic differentiation.

Method: Logarithmic Differentiation

Logarithmic differentiation is a useful technique when dealing with functions of the form $ y = [f(x)]^{g(x)} $. It involves taking the natural logarithm of both sides to simplify the expression before differentiating.

Step 1: Take the Natural Logarithm

Start with the given function:

$ y = x^{\tan(x)} $

Take the natural logarithm (ln) of both sides:

$ \ln(y) = \ln(x^{\tan(x)}) $

Using the logarithm power rule $ \ln(a^b) = b \ln(a) $, we get:

$ \ln(y) = \tan(x) \ln(x) $

Step 2: Differentiate Implicitly

Now, differentiate both sides of the equation with respect to $ x $. Remember to use the chain rule for $ \ln(y) $ (which gives $ \frac{1}{y} \frac{dy}{dx} $) and the product rule for the right side ($ \frac{d}{dx}(uv) = u'v + uv' $).

• Let $ u = \tan(x) $ and $ v = \ln(x) $.
• Then $ u' = \frac{d}{dx}(\tan(x)) = \sec^2(x) $.
• And $ v' = \frac{d}{dx}(\ln(x)) = \frac{1}{x} $.

Applying the product rule to $ \tan(x) \ln(x) $:

$ \frac{d}{dx}(\tan(x) \ln(x)) = (\sec^2(x))(\ln(x)) + (\tan(x))(\frac{1}{x}) $

So, the differentiation yields:

$ \frac{1}{y} \frac{dy}{dx} = \sec^2(x) \ln(x) + \frac{\tan(x)}{x} $

Step 3: Isolate $ \frac{dy}{dx} $

Multiply both sides by $ y $ to solve for $ \frac{dy}{dx} $:

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

Substitute the original expression for $ y $ back into the equation:

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

Evaluate the Derivative at $ x = \frac{\pi}{4} $

Now, we need to find the value of the derivative when $ x = \frac{\pi}{4} $. First, let's find the values of the trigonometric and logarithmic functions at this point:

• $ \tan(\frac{\pi}{4}) = 1 $
• $ \sec(\frac{\pi}{4}) = \frac{1}{\cos(\frac{\pi}{4})} = \frac{1}{1/\sqrt{2}} = \sqrt{2} $
• $ \sec^2(\frac{\pi}{4}) = (\sqrt{2})^2 = 2 $
• $ \ln(\frac{\pi}{4}) $ (This value remains as is)
• The base term $ x^{\tan(x)} $ becomes $ (\frac{\pi}{4})^{\tan(\frac{\pi}{4})} = (\frac{\pi}{4})^1 = \frac{\pi}{4} $

Substitute these values into the derivative formula:

$ \frac{dy}{dx} \bigg|_{x=\frac{\pi}{4}} = \left(\frac{\pi}{4}\right) \left( (2) \ln(\frac{\pi}{4}) + \frac{1}{\frac{\pi}{4}} \right) $

Simplify the expression:

$ \frac{dy}{dx} \bigg|_{x=\frac{\pi}{4}} = \frac{\pi}{4} \left( 2 \ln(\frac{\pi}{4}) + \frac{4}{\pi} \right) $

Distribute the $ \frac{\pi}{4} $:

$ \frac{dy}{dx} \bigg|_{x=\frac{\pi}{4}} = \left(\frac{\pi}{4} \cdot 2 \ln(\frac{\pi}{4})\right) + \left(\frac{\pi}{4} \cdot \frac{4}{\pi}\right) $

$ \frac{dy}{dx} \bigg|_{x=\frac{\pi}{4}} = \frac{\pi}{2} \ln(\frac{\pi}{4}) + 1 $

Conclusion and Option Matching

The calculated value of the derivative at $ x = \frac{\pi}{4} $ is $ \frac{\pi}{2} \ln(\frac{\pi}{4}) + 1 $.

Looking at the options, Option 3 is written as $ \frac{\pi}{4} \log(\frac{\pi}{4})^2 + 1 $. Assuming 'log' refers to the natural logarithm 'ln' and the notation implies $ \frac{\pi}{4} \log\left(\left(\frac{\pi}{4}\right)^2\right) + 1 $, we can use the logarithm property $ \log(a^b) = b \log(a) $:

$ \frac{\pi}{4} \log\left(\left(\frac{\pi}{4}\right)^2\right) + 1 = \frac{\pi}{4} \cdot 2 \log\left(\frac{\pi}{4}\right) + 1 = \frac{\pi}{2} \log\left(\frac{\pi}{4}\right) + 1 $

This result matches our calculation $ \frac{\pi}{2} \ln(\frac{\pi}{4}) + 1 $. Therefore, Option 3 correctly represents the value of the derivative.