Let

$f(x)=\frac{3x^4+4}{x^7+1}$

WITHOUT algebraic simplification, the Quotient Rule gives $f^{\text{'}}(x)tobe$:

$f^{\text{'}}(x)=$

$\frac{12x^3(x^7+1)-(3x^4+4)(7x^6)}{(x^7+1{)}^2}$

$12x^3(x^7+1)-(3x^4+4)(7x^6)$

$\frac{12x^3(x^7+1)+(3x^4+4)(7x^6)}{(x^7+1{)}^2}$

$\frac{12x^3}{7x^6}$

None $of$ the above.