y = (tanx)^(tanx) * tanx का अवकलन ज्ञात कीजिए।

Steps to Solve

1. Simplify the expression

First, simplify the given function $y = (\tan x)^{\tan x} \cdot \tan x$. Using exponent rules $a^m \cdot a^n = a^{m+n}$, we get:

$$y = (\tan x)^{\tan x + 1}$$

2. Apply Logarithmic Differentiation

Take the natural logarithm of both sides:

$$\ln y = \ln \left( (\tan x)^{\tan x + 1} \right)$$ $$\ln y = (\tan x + 1) \ln (\tan x)$$

3. Differentiate both sides with respect to $x$ Using the chain rule on the left side and the product rule on the right side:

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

4. Isolate $\frac{dy}{dx}$

Multiply both sides by $y$:

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

5. Substitute $y = (\tan x)^{\tan x + 1}$

$$\frac{dy}{dx} = (\tan x)^{\tan x + 1} \left[ \frac{(\tan x + 1)\sec^2 x}{\tan x} + \sec^2 x \ln (\tan x) \right]$$

6. Simplify (Optional)

We can further simplify the term inside the brackets: $$ \frac{dy}{dx} = (\tan x)^{\tan x + 1} \sec^2 x \left[ \frac{\tan x + 1}{\tan x} + \ln (\tan x) \right] $$ $$ \frac{dy}{dx} = (\tan x)^{\tan x + 1} \sec^2 x \left[ 1 + \frac{1}{\tan x} + \ln(\tan x) \right] $$ $$ \frac{dy}{dx} = (\tan x)^{\tan x + 1} \sec^2 x \left[ 1 + \cot x + \ln(\tan x) \right] $$