the kinematic equations are

$$x(t)=f(t)\quad\Rightarrow\\ v(t)=\frac{dx(t)}{dt}=\frac{df(t)}{dt}\\ a(t)=\frac{dv(t)}{dt}= \frac{d^2f(t)}{dt^2}$$

now if you want to obtain the acceleration $~a=a(x)~$ first you eliminate the parameter $~t~$ with the equation $~x=f(t)\quad\Rightarrow~t=g(x)~$ hence

$$a(x)=\frac{d^2f(t)}{dt^2}\bigg|_{t=g(x)}$$

but also with:

$$t=g(x)\quad,dt=\frac{\partial g(x)}{\partial x}\,dx\quad\Rightarrow\\ v(x)=\frac{dx}{dt}=\frac {1}{\frac{\partial g(x)}{\partial x}}\\ a(x)=\frac{dv}{dt}=\frac{dv(x)}{dx}\,\frac{dx}{dt}=\frac{dv(x)}{dx}\,v(x)$$

Example:

$$x=f(t)=c\,t^2\quad\Rightarrow\\ t=\frac {\sqrt{c\,x}}{c}=g(x)\quad,g'(x)=\frac{1}{2\sqrt{c\,x}}\\ v(x)=2\,\sqrt{c\,x}\quad,v'(x)=\frac{c}{\sqrt{c\,x}}\\ a(x)=\frac{c}{\sqrt{c\,x}}\,2\,\sqrt{c\,x}=2\,c $$

or
$$a(x)=\frac{d^2f(t)}{dt^2}\bigg|_{t=g(x)}=2\,c $$

the kinematic equations are

$$x(t)=f(t)\quad\Rightarrow\\ v(t)=\frac{dx(t)}{dt}=\frac{df(t)}{dt}\\ a(t)=\frac{dv(t)}{dt}= \frac{d^2f(t)}{dt^2}$$

now if you want to obtain the acceleration $~a=a(x)~$ first you eliminate the parameter $~t~$ with the equation $~x=f(t)\quad\Rightarrow~t=g(x)~$ hence

$$a(x)=\frac{d^2f(t)}{dt^2}\bigg|_{t=g(x)}$$

but also with:

$$t=g(x)\quad,dt=\frac{\partial g(x)}{\partial x}\,dx\quad\Rightarrow\\ v(x)=\frac{dx}{dt}=\frac {1}{\frac{\partial g(x)}{\partial x}}\\ a(x)=\frac{dv}{dt}=\frac{dv(x)}{dx}\,\frac{dx}{dt}=\frac{dv(x)}{dx}\,v(x)$$

Example:

$$x=f(t)=c\,t^2\quad\Rightarrow\\ t=\frac {\sqrt{c\,x}}{c}=g(x)\quad,g'(x)=\frac{1}{2\sqrt{c\,x}}\\ v(x)=2\,\sqrt{c\,x}\quad,v'(x)=\frac{c}{\sqrt{c\,x}}\\ a(x)=\frac{c}{\sqrt{c\,x}}\,2\,\sqrt{c\,x}=2\,c $$

or
$$a(x)=\frac{d^2f(t)}{dt^2}\bigg|_{t=g(x)}=2\,c $$

also $$a(x)=-{\frac {{\frac {d^{2}}{d{x}^{2}}}g \left( x \right) }{ \left( {\frac {d}{dx}}g \left( x \right) \right) ^{3}}} $$

the kinematic equations are

$$x(t)=f(t)\quad\Rightarrow\\ v(t)=\frac{dx(t)}{dt}=\frac{df(t)}{dt}\\ a(t)=\frac{dv(t)}{dt}= \frac{d^2f(t)}{dt^2}$$

now if you want to obtain the acceleration $~a=a(x)~$ first you eliminate the parameter $~t~$ with the equation $~x=f(t)\quad\Rightarrow~t=g(x)~$ hence

$$a(x)=\frac{d^2f(t)}{dt^2}\bigg|_{t=g(x)}$$

but also with:

$$t=g(x)\quad,dt=\frac{\partial g(x)}{\partial x}\,dx\quad\Rightarrow\\ v(x)=\frac{dx}{dt}=\frac {1}{\frac{\partial g(x)}{\partial x}}\\ a(x)=\frac{dv}{dt}=\frac{dv(x)}{dx}\,\frac{dx}{dt}=\frac{dv(x)}{dx}\,v(x)$$

Example:

$$x=f(t)=c\,t^2\quad\Rightarrow\\ t=\frac {\sqrt{c\,x}}{c}=g(x)\quad,g'(x)=\frac{1}{2\sqrt{c\,x}}\\ v(x)=2\,\sqrt{c\,x}\quad,v'(x)=\frac{c}{\sqrt{c\,x}}\\ a(x)=\frac{c}{\sqrt{c\,x}}\,2\,\sqrt{c\,x}=2\,c $$

and $$a(x)=\frac{d^2f(t)}{dt^2}\bigg|_{t=g(x)}=2\,c $$

the kinematic equations are

$$x(t)=f(t)\quad\Rightarrow\\ v(t)=\frac{dx(t)}{dt}=\frac{df(t)}{dt}\\ a(t)=\frac{dv(t)}{dt}= \frac{d^2f(t)}{dt^2}$$

now if you want to obtain the acceleration $~a=a(x)~$ first you eliminate the parameter $~t~$ with the equation $~x=f(t)\quad\Rightarrow~t=g(x)~$ hence

$$a(x)=\frac{d^2f(t)}{dt^2}\bigg|_{t=g(x)}$$

but also with:

$$t=g(x)\quad,dt=\frac{\partial g(x)}{\partial x}\,dx\quad\Rightarrow\\ v(x)=\frac{dx}{dt}=\frac {1}{\frac{\partial g(x)}{\partial x}}\\ a(x)=\frac{dv}{dt}=\frac{dv(x)}{dx}\,\frac{dx}{dt}=\frac{dv(x)}{dx}\,v(x)$$

Example:

$$x=f(t)=c\,t^2\quad\Rightarrow\\ t=\frac {\sqrt{c\,x}}{c}=g(x)\quad,g'(x)=\frac{1}{2\sqrt{c\,x}}\\ v(x)=2\,\sqrt{c\,x}\quad,v'(x)=\frac{c}{\sqrt{c\,x}}\\ a(x)=\frac{c}{\sqrt{c\,x}}\,2\,\sqrt{c\,x}=2\,c $$

or
$$a(x)=\frac{d^2f(t)}{dt^2}\bigg|_{t=g(x)}=2\,c $$