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  • <img alt="y=\frac{x^2}{(x-a)(x-b)} \text { then } \frac{d^n y}{d x^n}=" src="https://entrancecorner.oncodecogs.com/gif.latex?y%3D%5Cfrac%7Bx%5E2%7D%7B%28x-a%29%28x-b%29%7D%20%5Ctext%20%7B%20then%20

y=\frac{x^2}{(x-a)(x-b)} \text { then } \frac{d^n y}{d x^n}=

Option: 1

\frac{(-1)^{\mathrm{n}}\lfloor\mathrm{n}+1}{(\mathrm{a}-\mathrm{b})^{\mathrm{n}+1}}\left[\frac{\mathrm{a}^2}{(\mathrm{x}-\mathrm{a})^{\mathrm{n}}}-\frac{\mathrm{b}^2}{(\mathrm{x}-\mathrm{b})^{\mathrm{n}}}\right]


Option: 2

\frac{\lfloor n}{(a-b)}\left[(x-a)^n-(x-b)^n\right]


Option: 3

\frac{(-1)^{\mathrm{n}}\lfloor\mathrm{n}}{(\mathrm{a}-\mathrm{b})}\left[\frac{\mathrm{a}^2}{(\mathrm{x}-\mathrm{a})^{\mathrm{n}+1}}-\frac{\mathrm{b}^2}{(\mathrm{x}-\mathrm{b})^{\mathrm{n}+1}}\right]


Option: 4

\frac{(-1)^n\lfloor 2 n}{(a-b)^n}\left[\frac{a^2}{(x-a)^{2 n}}-\frac{b^2}{(x-b)^{2 n}}\right]


Answers (1)

best_answer

y=\frac{x^2}{(x-a)(x-b)}

Since the given fraction is not a proper one, therefore we should first divide the numerator by the denominator before resolving it into particle fractions. Here we observe orally that the quotient will be 1. so let


\begin{aligned} & \frac{x^2}{(x-a)(x-b)} \equiv 1+\frac{A}{x-a}+\frac{B}{x-b} \\ & \text { then } A=\frac{a^2}{a-b} \text { and } B=\frac{b^2}{b-a} \end{aligned}

hence,

y=1+\frac{a^2}{(a-b)(x-a)}-\frac{b^2}{(a-b)(x-b)}=1+\frac{a^2}{(a-b)(x-a)}-\frac{b^2}{(a-b)(x-b)}

Now differentiating both sides n times, we get

\mathrm{y}_{\mathrm{n}}=\frac{\mathrm{a}^2}{(\mathrm{a}-\mathrm{b})}(-1)^{\mathrm{n}} \mathrm{n} !(\mathrm{x}-\mathrm{a})^{-\mathrm{n}-1}-\frac{\mathrm{b}^2}{(\mathrm{a}-\mathrm{b})}(-1)^{\mathrm{n}} \mathrm{n} !(\mathrm{x}-\mathrm{b})^{-\mathrm{n}-1}=\frac{(-1)^{\mathrm{n}} \mathrm{n} !}{(\mathrm{a}-\mathrm{b})}\left[\frac{\mathrm{a}^2}{(\mathrm{x}-\mathrm{a})^{\mathrm{n}+1}}-\frac{\mathrm{b}^2}{(\mathrm{x}-\mathrm{b})^{\mathrm{n}+1}}\right]

 

Posted by

Rishi

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