17. If [tex]$y=\frac{(1-x)^{20}}{(1+z)^m}$[/tex], find [tex]$\frac{d y}{d x}$[/tex].

A. [tex]$y\left(\frac{2 a}{1+\pi}+\frac{25}{15}\right)$[/tex]
B. [tex][tex]$y\left(\frac{20}{1+\pi}+1 \frac{20}{1-\pi}\right)$[/tex][/tex]
C. [tex]$-y\left(\frac{21}{1-2}+\frac{2 n}{1+x}\right)$[/tex]
D. [tex]$-y\left(\frac{21}{1-x}+\frac{25}{1+x}\right)$[/tex]



Answer :

To find the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] for the given function [tex]\( y = \frac{(1-x)^{20}}{(1+z)^m} \)[/tex], we should proceed as follows:

### Step-by-Step Solution:

Given:
[tex]\[ y = \frac{(1-x)^{20}}{(1+z)^m} \][/tex]

We need to find [tex]\( \frac{dy}{dx} \)[/tex].

1. Identify the components to be differentiated:
- The numerator is [tex]\((1-x)^{20}\)[/tex].
- The denominator is [tex]\((1+z)^m\)[/tex] which is constant with respect to [tex]\( x \)[/tex].

2. Apply the power rule for differentiation to the numerator:
- The power rule states that the derivative of [tex]\( u^n \)[/tex] with respect to [tex]\( u \)[/tex] is [tex]\( n \cdot u^{n-1} \)[/tex].
- Here, [tex]\( u = 1 - x \)[/tex] and [tex]\( n = 20 \)[/tex].

3. Derivation of the numerator:
[tex]\[ \frac{d}{dx} (1 - x)^{20} = 20 (1 - x)^{19} \cdot \frac{d}{dx} (1 - x) \][/tex]
Given [tex]\(\frac{d}{dx} (1 - x) = -1\)[/tex]:
[tex]\[ \frac{d}{dx} (1 - x)^{20} = 20 (1 - x)^{19} \cdot (-1) = -20 (1 - x)^{19} \][/tex]

4. Consider the denominator [tex]\((1+z)^m\)[/tex]:
- Since it is constant with respect to [tex]\( x \)[/tex], it remains unchanged during the differentiation.

5. Combine the results:
[tex]\[ \frac{dy}{dx} = \frac{-20 (1-x)^{19}}{(1+z)^m} \][/tex]

Thus:
[tex]\[ \boxed{\frac{dy}{dx} = \frac{-20 (1-x)^{19}}{(1+z)^m}} \][/tex]

By comparing the above result with the provided options, the matching option is: [tex]\( \boxed{-y \left( \frac{21}{1-x} + \frac{25}{1+x} \right)} \)[/tex].

However, there seems to be a discrepancy because the correct result should match the form:

[tex]\[ -20 \frac{(1 - x)^{19}}{(1+z)^m} = y \left(-20 \frac{(1-x)^{19}}{(1-x)^{20}}\right) \][/tex]

Since:
[tex]\[ y = \frac{(1-x)^{20}}{(1+z)^m} \Rightarrow \frac{y}{(1-x)^{20}}=\frac{1}{1+z} \text{(which is not among given options)} \][/tex]

Nonetheless, note that none of the provided choices seems to accurately reflect the result of: [tex]\(-20 \frac{(1-x)^{19}}{(1+z)^m}\)[/tex]. Thus if we select the closest theoretical match, simplify the choices, it would potentially be option (B) if it was formulated clearly.

Given condition assignments might have inaccuracies in the understanding presented - it appears data misalignment with expected answers.

Hence correct independent result found:
[tex]\[ -20 \frac{(1-x)^{19}}{(1+z)^m} \][/tex]
is \boxed{-\frac{20 \cdot (1-x)^{19}}{(1+z)^m}}