# Some problem when solving the second derivative of graph expressed by parameter equation

For example, a graph expressed by this $$x=sin(t)$$ $$y=e^{t}+1$$ What my teacher told me to solve $$\frac{\mathrm{d^{2}} y}{{dx}^{2}}$$ is like: $$\frac{\mathrm{d^{2}} y}{{dx}^{2}}=\frac{\mathrm{d}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right ) }{\mathrm{d} x}=\frac{\mathrm{d}\left ( \frac{\mathrm{d} y}{\mathrm{d} t}\frac{\mathrm{d} t}{\mathrm{d} x} \right ) }{\mathrm{d} x}=\frac{\mathrm{d}\left ( \frac{e^{t}}{cos(t)} \right ) }{\mathrm{d} x}=\frac{\mathrm{d}\left ( \frac{e^{t}}{cos(t)} \right ) }{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} x}=\frac{e^{t}cos(t)+e^{t}sin(t)}{{cos(t)}^3}$$ I want to ask, why can't I do this question like this: $$\frac{\mathrm{d^{2}} y}{{dx}^{2}}=\frac{\mathrm{d^{2}} y}{{dt}^{2}}\frac{1}{(\frac{\mathrm{d} x}{\mathrm{d} t})^{2}}=\frac{e^{t}}{{cos(t)}^{2}}$$ why do I have to solve this question using the first way? And I also want to ask about the geometric meaning, just like the first derivative of a graph expressed by parameter equation is the slope of the tangent line, of the second derivative.

The followind identity you are using is not correct: $$\frac{\mathrm d^2 y}{\mathrm d x^2} = \frac{\mathrm d^2 y}{\mathrm d t^2} \frac{1}{\left(\frac{\mathrm d x}{\mathrm d t}\right)^2}.$$ Keep in mind that, although sometimes it may look like so, you cannot treat $$\mathrm dx$$ and $$\mathrm dy$$ as fraction numerators and denominators.

Your teacher provided the correct path to the solution, because you are supposed to think about how $$y$$ indirectly 'depends' on $$x$$ through $$t$$. Let's say you have $$t = f(x)$$, which is not known. Then $$y(x) = e^{f(x)} - 1$$. If you want to find the derivative with respect to $$x$$, you use the chain rule. The second derivative you are looking for is: $$\frac{\mathrm d^2 y}{\mathrm d x^2} = \frac{\mathrm d \left(\frac{\mathrm d y}{\mathrm d x}\right)}{\mathrm d x} = \frac{\mathrm d \left(\frac{\mathrm d y(f(x))}{\mathrm d t}\frac{\mathrm df(x)}{\mathrm dx} \right)}{\mathrm d x} = \dots$$ The rest of the derivation is as you were instructed.

• Just typed the complete answer , and Few seconds ago, This post was posted, It pretty much sums up What I wanted to say.. Nov 8, 2023 at 14:19
• It looks like that I shouldn't treat any derivative that upper than the first derivative as fraction, is that right? If it's right, please tell me why. And I actually questioned two questions, can you answer me the second one as well? Thanks a lot for unreserved help! Nov 8, 2023 at 14:52
• @QingGuang233 In fact, it's better to never treat a derivative as a fraction. The small changes fraction intuition is useful for students when they think about physical phenomena, but it is very misleading in mathematics. That is because the derivative at a point is a limit, not a ratio. This means that the derivative at all points, if it exists, is just a function that may, OR may not, approximate the small changes ratio. Nov 9, 2023 at 12:20

HINT:

For any $$t$$ parametrization with dotted derivatives $$\frac{d^2 y}{d x^2} = \frac{d \left(\frac{ dy}{dx}\right)}{dx}=\frac{d \left(\frac{\mathrm dy}{ d x}\right)/dt}{ d x/dt}$$

$$= \frac{d\left ( {\dot {y}} / {\dot x}\right)/dt}{\dot x}$$

$$=\dfrac{ \dot x\ddot y-\ddot x \dot y}{\dot x^3}$$ which you can find by individual $$x,y$$ derivatives.