Question

d\dx(1+x²+x⁴)\(1+x+x²)​

Answer 1

${\tt{\underline{Given:-}}}$

$\bullet \bf \dfrac{d}{dx}\bigg(\dfrac{1+x^2+x^4}{1+x+x^2}\bigg)=ax+b$

${\tt{\underline{Need\;to\;find:-}}}$

The respective values of a and b

${\tt{\underline{Answer:-}}}$

$\bullet \tt\dfrac{d}{dx}\bigg(\dfrac{1+x^2+x^4}{1+x+x^2}\bigg)$

$\tt \dfrac{d}{dx}\bigg(\dfrac{1+2x^2+x^4-x^2}{1+x+x^2}\bigg)=ax+b$

$\tt\dfrac{d}{dx}\bigg(\dfrac{(1+x^2)^2 - x^2}{1+x+x^2}\bigg)=ax+b$

$\tt\dfrac{d}{dx}\bigg(\dfrac{(1+x^2+x)(1+x^2-x)}{1+x+x^2}\bigg)=ax+b$

$\tt\dfrac{d}{dx}\bigg(\dfrac{\cancel{(1+x^2+x)}+(1+x^2-x)}{\cancel{(1+x^2+x)}}\bigg)=ax+b$

$\tt\dfrac{d}{dx}\bigg(1+x^2+x\bigg)=ax+b$

$\tt 2x-1=ax+b$

Here, we are having the same form of the equation as the given L.H.S

$\bf\underline{ 2=a\;\;\;\;\& \;\;\;\; -1=b}$

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