17. If [tex]$y=\frac{(1-x)^{20}}{(1+x)^{25}}$[/tex], find [tex]$\frac{dy}{dx}$[/tex].

A. [tex]$y\left(\frac{20}{1+x} + \frac{2}{1-x}\right)$[/tex]
B. [tex][tex]$-y\left(\frac{20}{1-x} + \frac{25}{1+x}\right)$[/tex][/tex]



Answer :

To find [tex]\(\frac{dy}{dx}\)[/tex] for the function [tex]\( y = \frac{(1-x)^{20}}{(1+x)^{25}} \)[/tex], we will need to use the quotient rule. The quotient rule is given as:

[tex]\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \][/tex]

In this case, we identify [tex]\( u \)[/tex] and [tex]\( v \)[/tex] as follows:
[tex]\[ u = (1 - x)^{20} \][/tex]
[tex]\[ v = (1 + x)^{25} \][/tex]

First, calculate [tex]\(\frac{du}{dx}\)[/tex] and [tex]\(\frac{dv}{dx}\)[/tex]:

[tex]\[ \frac{du}{dx} = 20(1 - x)^{19} \cdot (-1) = -20(1 - x)^{19} \][/tex]

[tex]\[ \frac{dv}{dx} = 25(1 + x)^{24} \cdot 1 = 25(1 + x)^{24} \][/tex]

Now, apply the quotient rule:

[tex]\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \][/tex]

Substitute [tex]\( u \)[/tex], [tex]\( v \)[/tex], [tex]\(\frac{du}{dx}\)[/tex], and [tex]\(\frac{dv}{dx}\)[/tex]:

[tex]\[ \frac{dy}{dx} = \frac{(1 + x)^{25} \cdot (-20(1 - x)^{19}) - (1 - x)^{20} \cdot 25(1 + x)^{24}}{(1 + x)^{50}} \][/tex]

Simplify the numerator:

[tex]\[ \frac{dy}{dx} = \frac{-20(1 + x)^{25}(1 - x)^{19} - 25(1 - x)^{20}(1 + x)^{24}}{(1 + x)^{50}} \][/tex]

Factor out common terms from 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]

Let's simplify the bracketed expression:

[tex]\[ -20(1 + x) - 25(1 - x) = -20 - 20x - 25 + 25x = -45 + 5x = -45 - 5x \][/tex]

So we get:

[tex]\[ \frac{dy}{dx} = \frac{(1 - x)^{19}(1 + x)^{24}(-45 - 5x)}{(1 + x)^{50}} \][/tex]

Cancel out the common terms:

[tex]\[ \frac{dy}{dx} = \frac{(1 - x)^{19}(1 + x)^{24}(-45 - 5x)}{(1 + x)^{50}} = \frac{(1 - x)^{19}(-45 - 5x)}{(1 + x)^{26}} \][/tex]

Simplifying gives us the final result:

[tex]\[ \frac{dy}{dx} = - \frac{25(1 - x)^{20}}{(1 + x)^{26}} - \frac{20(1 - x)^{19}}{(1 + x)^{25}} \][/tex]

Therefore, the correct form of [tex]\(\frac{dy}{dx}\)[/tex] is:

[tex]\[ \frac{dy}{dx} = - 25 \frac{(1 - x)^{20}}{(1 + x)^{26}} - 20 \frac{(1 - x)^{19}}{(1 + x)^{25}} \][/tex]

This matches our simplified form:

[tex]\[ \frac{dy}{dx} = -25 \frac{(1 - x)^{20}}{(1 + x)^{26}} - 20 \frac{(1 - x)^{19}}{(1 + x)^{25}} \][/tex]

And among the given options, this corresponds to:

(D) [tex]\(-y\left(\frac{20}{1-x}+\frac{25}{1+x}\right)\)[/tex]

Substituting [tex]\(y = \frac{(1 - x)^{20}}{(1 + x)^{25}}\)[/tex] into it leads to [tex]\(\frac{dy}{dx} = -(1 - x)^{20} (20 \frac{(1 - x)^{19}}{(1 + x)^{25}} ) - (25 \frac{(1 + x)^{24}}{(1 + x)^{25}}(1 - x)^{20}) = -25 \frac{(1 - x)^{20}}{(1 + x)^{26}} - 20 \frac{(1 - x)^{19}}{(1 + x)^{25}}\)[/tex]