Question

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

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

Answer 1

Answer: (C)

$\frac{-g\left(t\right)}{f\left(t\right)}$

$x=f\left(t\right)=a^{\ln \left(b^t\right)}=a^{t\ln b}$ --------- (1)
$y=g\left(t\right)=b^{-\ln \left(a^t\right)}={\left(b^{\ln a}\right)}^{-t}={\left(a^{\ln b}\right)}^{-t}=a^{-t\ln b}$
$\therefore y=g\left(t\right)=a^{\ln \left(b^{-t}\right)}=f(-t)$ -------- (2)
From equations $\left(1\right)$ and $\left(2\right)$,
$xy=1$
$\Rightarrow y=\frac{1}{x}$
$\therefore \frac{dy}{dx}=-\frac{1}{x^2}=-\frac{1}{f^2\left(t\right)}$
Also, $xy=1⟹-\frac{1}{f^2\left(t\right)}=-g^2\left(t\right)$
$\therefore \frac{dy}{dx}=-\frac{1}{x^2}=-\frac{y^2}{1}$
Also, $xy=1⟹\frac{dy}{dx}=-\frac{y}{x}=-\frac{g\left(t\right)}{f\left(t\right)}$