Question

question if (t(x)=\frac{0.04e^{x}}{ln(x)}) what is (t(x))? select the correct answer below: (t(x)=0.04\frac{ln x(e^{x})-(e^{x})(\frac{1}{x})}{(ln x)^{2}}) (t(x)=0.04\frac{(e^{x})(\frac{1}{x})-ln x(e^{x})}{(ln x)^{2}}) (t(x)=0.04\frac{ln x(xe^{x - 1})-(e^{x})(\frac{1}{x})}{(ln x)^{2}}) (t(x)=\frac{0.04e^{x}}{\frac{1}{x}})

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 0.04e^{x}$ and $v=\ln(x)$.

Step2: Find $u'$ and $v'$

$u'=0.04e^{x}$ (since the derivative of $e^{x}$ is $e^{x}$) and $v'=\frac{1}{x}$ (derivative of $\ln(x)$).

Step3: Apply quotient - rule

$t'(x)=0.04\frac{e^{x}\ln(x)-e^{x}\frac{1}{x}}{(\ln(x))^{2}}$

Answer:

$t'(x)=0.04\frac{\ln x(e^{x})-(e^{x})(\frac{1}{x})}{(\ln x)^{2}}$