Question

The derivative of $\frac{7x^2}{4e^x-x}$ will be

• A. $\frac{7(8e^x-4x{e}^x-x)}{(4e^x-x{)}^2}$
• B. $\frac{7x(8e^x-4x{e}^x-x)}{(4e^x-x{)}^2}$
• C. $\frac{7x(8e^x-4x{e}^x-1)}{(4e^x-x{)}^2}$
• D. $\text{None of these}$

Answer 1

The correct option is B $\frac{7x(8e^x-4x{e}^x-x)}{(4e^x-x{)}^2}$

It’s easy to figure out that the function is quotient of two functions $7x^{2}and4e^x-x$. So we will use the quotient rule which says

$\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{f^{\prime }\left(x\right)\times g\left(x\right)-g^{\prime }\left(x\right)\times f\left(x\right)}{\left(g\right(x){)}^2}$

$f\left(x\right)hereis7x^2,f^{\prime }\left(x\right)=14x$
$g\left(x\right)hereis4e^x-x,g^{\prime }\left(x\right)willbe4e^x-1$
Just substitute the terms in the formula and the derivative will be
$\frac{(4e^x-x)\left(f^{\prime }\right(x\left)\right)-7x^2\times g^{\prime }\left(x\right)}{\left(g\right(x){)}^2}=\frac{(4e^x-x)\times \left(14x\right)-\left(7x^2\right)\times (4e^x-1)}{(4e^x-x{)}^2}$

$=\frac{56x{e}^x-14x^2-14x^2-(28x^2{e}^x-7x^2)}{(4e^x-x{)}^2}$

$=\frac{56x{e}^x-14x^2-28x^2{e}^x+7x^2}{(4e^x-x{)}^2}$

$=\frac{7x(8e^x-4x{e}^x-x)}{(4e^x-x{)}^2}.$