d^2y/dy^2 expressed in t

Question

Given that $$ x = tln(4t) $$ $$ y = t^3 + 4t^2 $$ Find $ \frac{d^2y}{dx^2} $ in terms of t

For this question is it right for me to say $$ dx/dt = tln(4t)dt=1+ln(4t) $$ $$ dy/dt = t^3dt+4t^2dt = 3t^2+8t $$ So, $$ \frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{dy}{dx} = \frac{3t^2+8t}{1+ln(4t)} $$ Hence, $$ \frac{d^2y}{dx^2} = \frac{3t^2+8t}{1+ln(4t)} {dt} = \frac{2(4+3t)*ln(4t)-3t}{(1+ln4t)^2} $$ ? My answer did not match with the answer key's

For the record, the answer key's answer is $$ \frac{d^2y}{dx^2} = \frac{2(4+3t)*ln(4t)-3t}{(1+ln4t)^3} $$

Answer 1

$$\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy/dt}{dx/dt})\cdot \frac{1}{\color{green}{\frac{dx}{dt}}}$$

Looks fine, except you forgot the $\frac{dx}{dt}$ factor in the denominator.

Answer 2

How the heck did you jump from $\frac{dy}{dx}$ to $\frac{d^2y}{dx^2}$ just by putting a "dt" on the end?

It looks to me like you calculated $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ NOT $\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)$.

Answer 3

$$\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\displaystyle \frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\displaystyle \frac{{dx}}{{dt}}}}$$

$$\frac{{{d^2}y}}{{d{x^2}}} =\frac{\frac{3 t+(6 t+8) \log (4 t)}{(\log (4 t)+1)^2}}{1 + \log(4 t)}=\frac{3 t+(6 t+8) \log (4 t)}{(\log (4 t)+1)^3}$$