Q.Find $\dfrac{dy}{dx}$ if $x=\sqrt{a^2+m^2}$, $y=\log(a^2+m^2)$ (parameter $m$).

Sometimes both $x$ and $y$ are given not directly in terms of each other but through a third variable $t$ (the parameter): $x=f(t)$, $y=g(t)$. For example $x=a\cos t$, $y=a\sin t$ traces a circle $x^2+y^2=a^2$ as $t$ varies, without $y$ ever being written explicitly as a function of $x$. When $x$ and $y$ are both differentiable functions of the parameter $t$ with $\dfrac{dx}{dt}\ne0$, then $y$ is a differentiable function of $x$, and $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ — found by differentiating $y$ and $x$ separately with respect to $t$ and dividing. This follows from the increment definition: $\dfrac{\delta y}{\delta x}=\dfrac{\delta y/\delta t}{\delta x/\delta t}$, and taking the limit as $\delta t\to0$ (so $\delta x\to0$ too, since $x$ is continuous in $t$) gives the result. To evaluate $dy/dx$ at a particular value of the parameter, compute $dy/dt$ and $dx/dt$ as functions of $t$, form the ratio, and then substitute the given value of $t$ (or $\theta$) at the end.