To find the derivative of the function [tex]\( y = \frac{(1-x)^{20}}{(1+x)^{25}} \)[/tex] with respect to [tex]\( x \)[/tex], we need to apply the rules of differentiation. Let's break this down step by step.
### Step-by-Step Solution
1. Identify the Function to Differentiate:
We have [tex]\( y = \frac{(1-x)^{20}}{(1+x)^{25}} \)[/tex].
2. Apply the Quotient Rule:
The quotient rule states that if [tex]\( y = \frac{u(x)}{v(x)} \)[/tex], then:
[tex]\[
\frac{dy}{dx} = \frac{u'v - uv'}{v^2}
\][/tex]
Here, [tex]\( u(x) = (1-x)^{20} \)[/tex] and [tex]\( v(x) = (1+x)^{25} \)[/tex].
3. Differentiate [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
- For [tex]\( u(x) = (1-x)^{20} \)[/tex]:
[tex]\[
u'(x) = \frac{d}{dx}[(1-x)^{20}] = 20(1-x)^{19} \cdot (-1) = -20(1-x)^{19}
\][/tex]
- For [tex]\( v(x) = (1+x)^{25} \)[/tex]:
[tex]\[
v'(x) = \frac{d}{dx}[(1+x)^{25}] = 25(1+x)^{24}
\][/tex]
4. Substitute [tex]\( u(x), u'(x), v(x), \)[/tex] and [tex]\( v'(x) \)[/tex] into the Quotient Rule:
[tex]\[
\frac{dy}{dx} = \frac{(-20(1-x)^{19})(1+x)^{25} - (1-x)^{20}(25(1+x)^{24})}{(1+x)^{50}}
\][/tex]
5. Simplify the Expression:
Since the common denominator is [tex]\( (1+x)^{50} \)[/tex]:
[tex]\[
\frac{dy}{dx} = \frac{-20(1-x)^{19}(1+x)^{25} - 25(1-x)^{20}(1+x)^{24}}{(1+x)^{50}}
\][/tex]
Factor out the common term in the numerator:
[tex]\[
\frac{dy}{dx} = \frac{(1-x)^{19}(1+x)^{24} \left[ -20(1+x) - 25(1-x) \right]}{(1+x)^{50}}
\][/tex]
[tex]\[
= \frac{(1-x)^{19}(1+x)^{24} \left[ -20 - 20x - 25 + 25x \right]}{(1+x)^{50}}
\][/tex]
[tex]\[
= \frac{(1-x)^{19}(1+x)^{24} \left[ -45 + 5x \right]}{(1+x)^{50}}
\][/tex]
[tex]\[
= \frac{(1-x)^{19}(-45 + 5x)}{(1+x)^{26}}
\][/tex]
6. Further Simplify the Fraction:
Distribute the 5:
[tex]\[
= \frac{5(9 - x)(1-x)^{19}}{(1+x)^{26}}
\][/tex]
This matches the simplified form [tex]\( 5 \cdot \frac{(9-x) (1-x)^{19}}{(1+x)^{26}} \)[/tex].
Thus, the derivative of [tex]\( y = \frac{(1-x)^{20}}{(1+x)^{25}} \)[/tex] is:
[tex]\[
\frac{dy}{dx} = \frac{-25 (1-x)^{20}}{(1+x)^{26}} - \frac{20 (1-x)^{19}}{(1+x)^{25}}
\][/tex]
which simplifies to:
[tex]\[
\frac{dy}{dx} = 5 \cdot \frac{(9-x) (1-x)^{19}}{(1+x)^{26}}
\][/tex]
Therefore, the correct answer matches:
[tex]\[
\boxed{-y\left(\frac{20}{1-x}+\frac{25}{1+x}\right)}
\][/tex]
