Question

If $y=(\tan x)^{x}$, then $\frac{1}{y}\frac{dy}{dx}=$ ________.

• A. $\log(\tan x)+2x \csc(2x)$
• B. $\log(\tan x)+x \csc(2x)$
• C. $x \log(\tan x)+2x \csc(2x)$
• D. $x \log(\tan x)+x^{2} \csc(2x)$
• E. $\log(\tan x)+\frac{x}{2} \csc(2x)$

Hint:

Use log differentiation for functions of the form $f(x)^{g(x)}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
We need to differentiate a function of the form \(y = [f(x)]^{g(x)}\). This type of function requires logarithmic differentiation. The question asks for \(\frac{1}{y}\frac{dy}{dx}\), which is precisely the derivative of \(\log y\).
Step 2: Key Formula or Approach
1. Take the natural logarithm of both sides of the equation \(y = (\tan x)^x\).
2. Use the logarithm property \(\log(a^b) = b\log(a)\) to simplify the right side.
3. Differentiate both sides with respect to \(x\). Use the chain rule for the left side (\(\frac{d}{dx}(\log y) = \frac{1}{y}\frac{dy}{dx}\)) and the product rule for the right side.
4. Simplify the resulting expression.
Step 3: Detailed Explanation
1. Take the natural logarithm.
Given \(y = (\tan x)^x\).
\[ \log y = \log((\tan x)^x) \] \[ \log y = x \log(\tan x) \] 2. Differentiate both sides with respect to x.
\[ \frac{d}{dx}(\log y) = \frac{d}{dx}(x \log(\tan x)) \] The left side is:
\[ \frac{1}{y} \frac{dy}{dx} \] For the right side, we use the product rule \((uv)' = u'v + uv'\) with \(u=x\) and \(v = \log(\tan x)\).
- \(u' = \frac{d}{dx}(x) = 1\).
- \(v' = \frac{d}{dx}(\log(\tan x))\). Using the chain rule, this is \(\frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x) = \frac{1}{\tan x} \cdot \sec^2 x\).
So, the derivative of the right side is:
\[ (1) \cdot \log(\tan x) + x \cdot \left(\frac{1}{\tan x} \sec^2 x\right) \] 3. Simplify the expression.
\[ \frac{1}{y} \frac{dy}{dx} = \log(\tan x) + x \frac{\sec^2 x}{\tan x} \] Let's simplify the second term:
\[ x \frac{\sec^2 x}{\tan x} = x \frac{1/\cos^2 x}{\sin x / \cos x} = x \frac{1}{\cos^2 x} \cdot \frac{\cos x}{\sin x} = x \frac{1}{\sin x \cos x} \] Now, use the double angle identity \(\sin(2x) = 2\sin x \cos x\), which implies \(\sin x \cos x = \frac{1}{2}\sin(2x)\).
\[ x \frac{1}{\sin x \cos x} = x \frac{1}{\frac{1}{2}\sin(2x)} = \frac{2x}{\sin(2x)} = 2x \csc(2x) \] 4. Combine the parts.
Substituting this back, we get:
\[ \frac{1}{y} \frac{dy}{dx} = \log(\tan x) + 2x \csc(2x) \] Step 4: Final Answer
The expression \(\frac{1}{y}\frac{dy}{dx}\) is equal to \(\log(\tan x) + 2x \csc(2x)\).