Q.6 (a) Find n^th derivative of y=x/1+3x+2x² .

1+3x+2x²=(1+x)(1+2x) . Partial fractions: x/(1+x)(1+2x)=frac A1+x+frac B1+2x⇒ x=A(1+2x)+B(1+x) . x=-1:\ -1=-A⇒ A=1 ; x=-1/2:\ -1/2=B/2⇒ B=-1 . y=frac11+x-frac11+2x (check x=1 : 1/2-1/3=1/6 ✓). Using Dⁿ(frac1ax+b)=frac(-1)ⁿn! aⁿ(ax+b)ⁿ⁺¹ : y_n=(-1)ⁿ n![frac1(1+x)ⁿ⁺¹-frac2ⁿ(1+2x)ⁿ⁺¹]

Find \(n^{th}\) derivative of \(y=\frac{x}{1+3x+2x^2}\).

Q.6 (a) Find \(n^{th}\) derivative of \(y=\frac{x}{1+3x+2x^2}\).

Doubtly Asked 25 Oct 2026 mu engineering Save WhatsApp Report

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\(1+3x+2x^2=(1+x)(1+2x)\). Partial fractions: \(\frac{x}{(1+x)(1+2x)}=\frac A{1+x}+\frac B{1+2x}\Rightarrow x=A(1+2x)+B(1+x)\).

\(x=-1:\ -1=-A\Rightarrow A=1\); \(x=-\frac12:\ -\frac12=\frac B2\Rightarrow B=-1\).

\(y=\frac1{1+x}-\frac1{1+2x}\) (check \(x=1\): \(\frac12-\frac13=\frac16\) ✓).

Using \(D^n\left(\frac1{ax+b}\right)=\frac{(-1)^nn!\,a^n}{(ax+b)^{n+1}}\):

\(y_n=(-1)^n\,n!\left[\frac1{(1+x)^{n+1}}-\frac{2^n}{(1+2x)^{n+1}}\right]\)

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

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Find \(n^{th}\) derivative of \(y=\frac{x}{1+3x+2x^2}\).

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