Question

$\bold{Find \: the \: derivative \: \frac{d}{dx}}$

$\begin{gathered} \\ \bf {x}^{2} \dfrac{ {d}^{2}y}{ {dx}^{2} } + 3x \dfrac{dy}{dx} + y = \dfrac{1}{ {(1 - x)}^{2} } \\ \end{gathered}$

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Answer 1

Apply basic rules of exponents.

Rewrite $\bold{\frac{1}{(1 - x)^{2}}}$ as $\: \bold{(( 1 - x) {}^{2} ) {}^{ - 1}}$

Multiply the exponents in $\bold{(( 1 - x) {}^{2} ) {}^{ - 1}}$

Apply the power rule and multiply exponents, $(a^m)^n$

$\frac{d}{dx}[ (1 - x) {}^{2 \times ( - 1)} ]$

Multiply 2 by -1

$\frac{d}{dx}[ (1 - x) {}^{ - 2} ]$

Differentiate using the chain rule, which states that $\frac{d}{dx}[f(g(x))$ is f'(g(x))g'(x) where f(x)=$x^-2$ and g(x)=1−x

To apply the Chain Rule, set u as 1-x.

$\frac{d}{du}[{u}^{ - 2}] \frac{d}{dx}[1 - x]$

Differentiate using the Power Rule which states that $\frac{d}{dx}[u {}^{ n} ]$ where n=-2

$- 2u {}^{ - 3} \frac{d}{dx}[1 - x]$

Replace all occurrences of u with 1-x

$- 2(1 - x) {}^{ - 3} \frac{d}{dx} [1 - x]$

Differentiate

By the Sum Rule, the derivative of 1-x with respect to x is $\frac{d}{dx}[1] + \frac{d}{dx} [ - x]$

$- 2(1 - x) {}^{ - 3} (\frac{d}{dx}[1] + \frac{d}{dx} [ - x])$

Since 1 is constant with respect to x, the derivative of 1 with respect to x is 0.

$- 2(1 - x) {}^{ - 3} (0 + \frac{d}{dx} [ - x])$

Add 0 and $\frac{d}{dx}$ [-x]

$- 2(1 - x) {}^{ - 3} \frac{d}{dx} [ - x]$

Since −1 is constant with respect to x, the derivative of −x with respect to x is -d/dx [x]

Differentiate using the Power Rule which states that $\frac{d}{dx} [ {x}^{n} ] \: is \: nx {}^{n - 1}$

where n is 1

$2(1 - x) {}^{ - 3}$

Rewrite the expression using the negative exponent rule $b^{-n}$=$\frac{1}{b^n}$

$2 \frac{1}{(1 - x) {}^{3} }$

Simplify

$Combine \: 2 \: and \: \frac{1}{(1 - x) {}^{3} } \\ \frac{2}{(1 - x) {}^{3} }$

Reorder terms

$\frac{2}{( - x + 1) {}^{3} }$