Calculate third derivative

Question

I need to calculate $\dfrac{d^3y}{dx^3}$, where $y = t^3$ and $x = e^{-t}$, basically I don't get how to use n-th differential formula when we also have x as function, could someone explain me this ?

Answer 1

Hint

We have that: $$\frac {dy}{dt}=3t^2$$ $$x=e^{-t} \implies x'=\frac {dx}{dt}=-e^{-t}$$ $$\frac {dy}{dx}=\frac {dy}{dt}\frac {dt}{dx}=\frac 1 {-e^{-t}}\frac {dy}{dt} =-{3t^2} {e^{t}} $$ $$\frac {d^2y}{dx^2}=-\frac {d(3t^2e^{t})}{dt}\frac {dt}{dx}=3te^{2t}(t+2) $$ $$.....$$ I let you finish..

Answer 2

HINT

We have that

• $x = e^{-t}\implies t=-\ln x\implies y=-\ln^3 x$