Question

If $x=f\left(t\right)$ and $y=g\left(t\right)$ are differentiable functions of $t$ then $\frac{d^{2}y}{d{x}^2}$ is

• A. $\frac{f^{\prime }\left(t\right).g"\left(t\right)-g^{\prime }\left(t\right).f"\left(t\right)}{\left[f^{\prime }\right(t){]}^3}$
• B. $\frac{f^{\prime }\left(t\right).g"\left(t\right)-g^{\prime }\left(t\right).f"\left(t\right)}{\left[f^{\prime }\right(t){]}^2}$
• C. $\frac{g^{\prime }\left(t\right).f"\left(t\right)-f^{\prime }\left(t\right).g"\left(t\right)}{\left[f^{\prime }\right(t){]}^3}$
• D. $\frac{g^{\prime }\left(t\right).f"\left(t\right)+f^{\prime }\left(t\right).g"\left(t\right)}{\left[f^{\prime }\right(t){]}^3}$

Answer 1

Answer: (A)

$\frac{f^{\prime }\left(t\right).g"\left(t\right)-g^{\prime }\left(t\right).f"\left(t\right)}{\left[f^{\prime }\right(t){]}^3}$

$x=f\left(t\right)$,$y=g\left(t\right)$

$\frac{dx}{dt}=f^{\prime }\left(t\right)$

$\frac{dy}{dt}=g^{\prime }\left(t\right)$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{g^{\prime }\left(t\right)}{f^{\prime }\left(t\right)}$

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

$=\frac{f^{\prime }\left(t\right).g"\left(t\right)-g^{\prime }\left(t\right).f"\left(t\right)}{\left(f^{\prime }\right(t){)}^2}.\frac{dt}{dx}$

$=\frac{f^{\prime }\left(t\right).g"\left(t\right)-g^{\prime }\left(t\right).f"\left(t\right)}{\left(f^{\prime }\right(t){)}^2}.\frac{1}{f^{\prime }\left(t\right)}$

$=\frac{f^{\prime }\left(t\right).g"\left(t\right)-g^{\prime }\left(t\right).f"\left(t\right)}{\left[f^{\prime }\right(t){]}^3}$