Question

If f(x) $=\frac{p{x}^2+qx+r}{ax+b}$, then f '(x)$=$

• A. $x^2+x$
• B. $\frac{ap{x}^2+2bpx+bq-ar}{(ax+b{)}^2}$
• C. $ap{x}^2+2bpx+pq-ar$
• D. None

Answer 1

Answer: (B)

$\frac{ap{x}^2+2bpx+bq-ar}{(ax+b{)}^2}$

$Givenf\left(x\right)=\frac{p{x}^2+qx+r}{ax+b}Differentiatew.r.tx{f}^{\prime }\left(x\right)=\frac{(2px+q)(ax+b)-(p{x}^2+qx+r)}{(ax+b{)}^2}=\frac{(2ap{x}^2+2bpx+aqx+bq)-(ap{x}^2+aqx+ar)}{(ax+b{)}^2}Byfurthersimplification=\frac{ap{x}^2+2bpx+bq-ar}{(ax+b{)}^2}HencethecorrectoptionisOPTIONB$