You can take the derivative $\frac{dy}{dx}$ of any relation involving $y$ and $x$ in order to find the slope.
As an example, consider the relation that describes a circle of radius $R$: \[ x^2 + y^2 = R^2. \] We are given a point $P=(x_P,y_P)$ that lies on the circle and asked to find the slope of the circle at that point. We begin by taking the implicit derivative of the relation that describes the circle \[ \begin{align*} \big[\ x^2 \ + \ \ y^2 \ \ \ &= R^2 \big]' \nl 2x + 2y\frac{dy}{dx} &= 0. \end{align*} \] After rearranging the terms a bit, we find that the slope of the circle at point $P=(x_P,y_P)$ is given by \[ \frac{dy}{dx} = - \frac{x_P}{y_P}. \] You can check that the slope predicted at $P=(0,R)$ is $0$ (the circle is flat at the top) and that at $P=(R,0)$, the slope is infinite since the tangent to the circle is vertical.
As you can see there is nothing fancy going on, we just went through and took the derivative of each term in the equation, and then used the result to isolate the slope $\frac{dy}{dx}$. In particular we didn't have to explicitly find the $y$ as a function of $x$, (which is actually possible $y=f(x)=\pm \sqrt{R^2-x^2}$). We took the derivative directly on expression describing the circle treating implicitly $y$ is a function of $x$.
expression with respect to the variable $x$.
In corporate America, a man's ego $E$ is related to his salary $S$ by the following equation: \[ E^2 = S^3. \] Assume that both $E$ and $S$ are implicitly functions of time. What is the rate of change of the ego of dirty Joe the insurance analyst, when he is making 60k and his salary is increasing at a rate of 5k per year?
We are told that $\frac{dS}{dt}=5000$ and we are asked to find $\frac{dE}{dt}$ when $S=60000$. To do this we first take the implicit derivative of the salary-to-ego relation as follows: \[ 2E \frac{dE}{dt} = 3 S^2 \frac{dS}{dt}. \] We are interested in the point where $S=60000$. To find Joe's ego at that point, we use the original relation $ E^2 = S^3$ and solving for $E$ we find $E=\sqrt{ 60000^3}= 14696938.46$ ego points. Substituting all these values into the derivative of the relation we find that \[ 2(14696938.46)\frac{dE}{dt} = 3 (60 000)^2 (5000). \] Joe's ego is growing at $\frac{dE}{dt}=\frac{3 (60 000)^2 (5000)}{2(14696938.46)} = 1837117.31$ ego points per year. Yey ego points! I wonder what you can redeem these for…
You want to calculate the precision associated with your measurement of the kinetic energy of some particle. Recall that the formula for the kinetic energy is $K = \frac{1}{2}m v^2$. Your measurement of the mass $m$ has precision 3%, and the measurement of the velocity has precision 2%. What will be the precision of your calculation for the kinetic energy?
The precision of any measurement is defined as ratio of the error divided by the quantity itself. We report our measurement of some quantity $Q$ as $Q \pm dQ$. The relative error (in percent) is $\frac{dQ}{Q}$.
In this case we want to find $\frac{dK}{K}$ and we are told $\frac{dm}{m}=0.03$ and $\frac{dv}{v}=0.02$. We proceed by calculus, taking the implicit derivative of the expression for kinetic energy: \[ dK = d\left( \frac{1}{2}m v^2 \right) = \frac{1}{2}(dm) v^2 + m v (dv), \] where we used the product rule of derivatives. In order to obtain ratios, we divide both sides by $K$ to obtain: \[ \frac{dK}{K} = \frac{\frac{1}{2}dm v^2 \ + \ m v dv}{ \frac{1}{2}m v^2 } = \frac{dm}{m} + 2 \frac{dv}{v}, \] so the precision of the kinetic energy measurement is $\frac{dK}{K}=0.03+ 2(0.02)=0.07$ or 7%.
The above examples illustrate that derivative rules are applicable very generally. Indeed we can talk about the differential or “change in” $dQ$ for any quantity $Q$.