If ` y=f(x) = ((tan^m x)/(tan^n x)) ^(m+n) .((tan^n x)/(tan^p x))((tan^p x)/(tan^m x)) ` then ` (dy)/(dx)`=

Options

A. 0
B. 1
C. `tan^(m+n+p)`x
D. None

Answer

To find the derivative \( \frac{dy}{dx} \) of the function given by

\[ y = f(x) = \left(\frac{\tan^m x}{\tan^n x}\right)^{m+n} \cdot \left(\frac{\tan^n x}{\tan^p x}\right) \cdot \left(\frac{\tan^p x}{\tan^m x}\right), \]

we will simplify the expression step by step before differentiating.

### Step 1: Simplify the expression

We can rewrite the function \( y \) as follows:

\[ y = \left(\tan^m x \cdot \tan^{-n} x\right)^{m+n} \cdot \left(\tan^n x \cdot \tan^{-p} x\right) \cdot \left(\tan^p x \cdot \tan^{-m} x\right). \]

Using the property of exponents \( a^m \cdot a^n = a^{m+n} \), we can combine the terms:

\[ y = \tan^{m(m+n)} x \cdot \tan^{-n(m+n)} x \cdot \tan^{n-p} x \cdot \tan^{p-m} x. \]

Now, we can combine the powers of \( \tan x \):

\[ y = \tan^{m(m+n) - n(m+n) + (n-p) + (p-m)} x. \]

### Step 2: Combine the exponents

Now, let's simplify the exponent:

\[ m(m+n) - n(m+n) + (n-p) + (p-m) = m^2 + mn - nm - n^2 + n - p + p - m. \]

This simplifies to:

\[ m^2 - n^2 + n - m. \]

So we have:

\[ y = \tan^{m^2 - n^2 + n - m} x. \]

### Step 3: Differentiate using the chain rule

Now we will differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx} \left(\tan^{m^2 - n^2 + n - m} x\right). \]

Using the chain rule, we have:

\[ \frac{dy}{dx} = (m^2 - n^2 + n - m) \tan^{m^2 - n^2 + n - m - 1} x \cdot \sec^2 x. \]

### Final Result

Thus, the derivative is:

\[ \frac{dy}{dx} = (m^2 - n^2 + n - m) \tan^{m^2 - n^2 + n - m - 1} x \cdot \sec^2 x. \]

---