Question

what is $r(x)$ if $r(x)=\frac{70x^{5}}{2^{x}}$? select the correct answer below: $r(x)=\frac{70(5x^{4})}{(2^{x})(ln 2)}$, $r(x)=\frac{70(2^{x})(5x^{4}) - 70(x^{5})(x)(2^{x - 1})}{2^{2x}}$, $r(x)=\frac{70(x^{5})(2^{x})(ln 2)-70(2^{x})(5x^{4})}{2^{2x}}$, $r(x)=\frac{70(2^{x})(5x^{4})-70(x^{5})(2^{x})(ln 2)}{2^{2x}}$

Explanation:

Step1: Apply quotient - rule

$(\frac{u}{v})'=\frac{u'v - uv'}{v^{2}}$, where $u = 70x^{5}$, $u'=70\times5x^{4}$, $v = 2^{x}$, $v'=2^{x}\ln2$.

Step2: Substitute values

$r'(x)=\frac{70(5x^{4})\times2^{x}-70(x^{5})\times2^{x}\ln2}{(2^{x})^{2}}=\frac{70(2^{x})(5x^{4})-70(x^{5})(2^{x})(\ln2)}{2^{2x}}$

Answer:

$r'(x)=\frac{70(2^{x})(5x^{4})-70(x^{5})(2^{x})(\ln2)}{2^{2x}}$