Question

A curve is represented parametrically by the equations $x=f\left(t\right)=a^{\ln \left(b^t\right)}$ and $y=g\left(t\right)=b^{-\ln \left(a^t\right)};a,b>0$ and $a\ne 1,b\ne 1$ where $t\in R$.

Which of the following is not a correct expression for $\frac{dy}{dx}?$

• A. $\frac{-1}{f(t{)}^2}$
• B. $-\left(g\right(t){)}^2$
• C. $\frac{-g\left(t\right)}{f\left(t\right)}$
• D. $\frac{-f\left(t\right)}{g\left(t\right)}$

Answer 1

The correct option is D $\frac{-f\left(t\right)}{g\left(t\right)}$

$x=f\left(t\right)=a^{\ln \left(b^t\right)}\Rightarrow x=a^{t\ln b}={\left(a^{\ln b}\right)}^t$
$y=g\left(t\right)=b^{-\ln \left(a^t\right)}\Rightarrow y=b^{-t\ln a}={\left(b^{\ln a}\right)}^{-t}\Rightarrow y={\left(a^{\ln b}\right)}^{-t}\Rightarrow y=a^{-t\ln b}=\frac{1}{x}\Rightarrow g\left(t\right)=f(-t)\Rightarrow xy=1\Rightarrow x\frac{dy}{dx}+y=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}=\frac{-f\left(t\right)}{g\left(t\right)}$