We have defined the derivative of $$y=f(x)$$ as the limit of $$\dfrac{\Delta y}{\Delta x}$$ and we have denoted the derivative by $$f'(x)$$ or $$\dfrac{dy}{dx}$$. The fractional form of $$dy/dx$$ has been suggesting that it is the limit of a fraction. So far we have made no attempt to treat $$dy/dx$$ as a fraction. In this section, we define $$dx$$ and $$dy$$ as separate quantities such that their ratio, when $$dx\neq0$$, will be $$f'(x)$$.

Definition. Suppose the function $$y=f(x)$$ is differentiable at the point $$x$$ (i.e. $$f'(x)$$ exits). Let $$dx$$ denote another independent variable which can take any value whatsoever $-\infty<dx<+\infty.$ $$dx$$ is called the “differential of $$x$$.”

The differential of $$f$$ is a function of two independent variables $$x$$ and $$dx$$ and its value is $$f'(x)dx$$: $df(x,dx)=f'(x)dx.$ It is customary to represent the value of the differential of $$f$$ for assigned values of $$x$$ and $$dx$$ by $$dy$$: $dy=f'(x)dx.\tag{i}$ $$dy$$ is called the differential of $$y$$. We may also simply write $$df=f'(x)dx$$.

• Similar to $$\Delta x$$ and $$\Delta y$$, $$dx$$ and $$dy$$ are single entities, not the product of $$d$$ and $$x$$ or $$d$$ and $$y$$.
Example 1

Given $$y=f(x)=x^{7}$$, find $$dy$$.

Solution 1

$f(x)=x^{7}\Rightarrow f'(x)=7x^{6}.$ Therefore, \begin{aligned} dy & =f'(x)\,dx\\ & =7x^{6}\:dx.\end{aligned}

Example 2

If $$y=f(x)=\sqrt{x}$$, calculate the value of $$dy$$ for $$x=4$$ and $$dx=0.1$$.

Solution 2

Here $$f'(x)=\dfrac{1}{2\sqrt{x}}$$, so $dy=\frac{1}{2\sqrt{x}}dx.$ Evaluating $$dy$$ for $$x=4$$ and $$dx=0.1$$, we obtain $dy=\left(\frac{1}{2\sqrt{4}}\right)(0.1)=(0.5)(0.1)=0.05.$

Example 3

Given $$v=f(u)=u\ln u$$, compute the differential $$dv$$.

Solution 3

In this example, the independent variable is $$u$$ and the dependent variable is $$v$$. \begin{aligned} f(u)=u\ln u\Rightarrow f'(u)=\frac{dv}{du} & =\ln u+u\cdot\frac{1}{u}\\ & =\ln u+1.\end{aligned} Therefore, $dv=f'(u)du=(\ln u+1)du.$

• If $$dx=0$$, then it follows from Formula (i) that $$dy=0$$.

• If $$dx\neq0$$, then we can divide (i) by $$dx$$ and obtain $\frac{dy}{dx}=\frac{\text{differential of }y\text{ in terms of }x\text{ and }dx}{\text{differential of }x}=\frac{f'(x)dx}{dx}=f'(x).\tag{ii}$ Therefore, following the present definition of the differentials $$dx$$ and $$dy$$, the derivative $$dy/dx=f'(x)$$ can be regarded as the ordinary quotient of $$dy$$ and $$dx$$. This result is, however, not surprising, because we planned to define $$dy$$ such that (ii) would be true.

• We should emphasize that $$dx$$ and $$dy$$ are NOT the limits of $$\Delta x$$ and $$\Delta y$$ as $$\Delta x\to0$$, because the limits of $$\Delta x$$ and $$\Delta y$$ are both zero.

### Geometric Interpretation and Linear Approximation

For a fixed value of $$x$$, say $$x=x_{0}$$, $$dy$$ is a dependent variable whose value is proportional to the independent variable $$dx$$. The proportionality coefficient is $$f'(x_{0})$$. Let us introduce a new rectangular coordinate system whose axes are labeled $$dx$$ and $$dy$$. In this coordinate system, the graph of the linear function given by $$dy=f'(x_{0})dx$$ is a straight line with slope $$f'(x_{0})$$ that passes through the origin (Figure 1(a)). If we place the origin of the $$dx$$$$dy$$ coordinate system at $$\left(x_{0},f(x_{0})\right)$$ (see Figure 1(b)), the graph of $$dy=f'(x_{0})dx$$ will coincide with the tangent line to the curve $$y=f(x)$$ at $$\left(x_{0},f(x_{0})\right)$$. In other words, the graph of the linear approximation $$y=f(x_{0})+f'(x_{0})(x-x_{0})$$ in the $$xy$$-plane becomes the graph of the linear function $$dy=f'(x_{0})dx$$ in the $$dx$$$$dy$$ plane.

Figure 1. (a) The graph of the function $$dy=f'(x_{0})dx$$ is a line with slope $$f'(x_{0})$$ passing through the origin in the $$dx$$$$dy$$ coordinate system. (b) If the $$dx$$$$dy$$ origin coincides with the point $$\left(x_{0},f(x_{0})\right)$$ and the $$dx$$ and $$dy$$ axes are parallel to the $$x$$ and $$y$$ axes, respectively, the graph of $$dy=f'(x_{0})dx$$ in the $$dx$$$$dy$$ plane coincides with the tangent line to the curve $$y=f(x)$$ at $$\left(x_{0},f(x_{0})\right)$$.

• A difference between $$\Delta x$$ and $$dx$$ is worth mentioning. The change in $$x$$ (or the increment of $$x$$), $$\Delta x$$, has to be so small such that $$x_{0}+\Delta x$$ will lie in the domain of $$f$$; however, $$dx$$ can be any real number, because the domain of the linear function $$dy=f'(x_{0})dx$$ is the entire set of real numbers $$\mathbb{R}$$.

Notice that $$dx$$ and $$dy$$ are not technically “infinitely small” quantities or “infinitesimals.” $$dx$$ can be any number, and for a large value of $$dx$$, $$dy$$ can be numerically very large too. However, in many applications, it is often useful to think of differentials as very small quantities. There are several good reasons for this way of thinking of differentials. One such reason is related to linear approximation. If $$\Delta x$$ is small and $$dx=\Delta x$$, then $$dy$$ is a good approximation for $$\Delta y$$ (see Figure 2): $\Delta y\approx dy$ or $\underbrace{f(x_{0}+dx)-f(x_{0})}_{\Delta y}\approx\underbrace{f'(x_{0})dx.}_{dy}$ This is exactly the linear approximation of $$f$$ (that we learned in the previous section) using the notation of differentials.

Example 4

Use differentials to approximate $$\sqrt[3]{65}$$.

Solution 4

When $$x=64=4^{3}$$, the evaluation of $$\sqrt[3]{x}$$ is easy, so we take $$f(x)=\sqrt[3]{x}$$ and $$dx=2$$. Because $$dy=f'(x)dx=\frac{1}{3}x^{-2/3}dx$$, for $$x=64$$ and $$dx=2$$, we have: $dy=\frac{1}{3(64^{2/3})}\cdot2=\frac{2}{3\times4^{2}}=\frac{1}{24}.$ Therefore $f(65)\approx f(64)+dy=4+\frac{1}{24}\approx4.04167.$ The actual value of $$\sqrt[3]{65}$$ to five decimals is $$4.04127$$, which shows that our approximation by differentials is accurate to three decimals places, although $$dx$$ is not very small.

Example 5

Use differentials to estimate $$4(1.001)^{5}-3(1.001)^{3}+2.$$

Solution 5

When $$x=1$$, the evaluation of $$f(x)=4x^{5}-3x^{3}+2$$ is easy $f(1)=4-3+2=3,$ so we take $$f(x)=4x^{4}-3x^{3}+2$$ and $$dx=0.001$$. Since $$dy=f'(x)dx=\left(20x^{4}-9x^{2}\right)dx$$, for $$x=1$$ and $$dx=0.001$$, we have: $dy=(20-9)(0.001)=11\times0.001=0.011.$ Therefore $f(1.001)\approx f(1)+dy=3+0.011=3.011.$ The actual value of $$f(1.001)$$ to five decimals is 3.01103.

If the true value of a quantity is $$Q_{t}$$ and the approximated value of the quantity (from a measurement or calculation) is $$Q_{a}$$ then $\text{true error}=\left|Q_{t}-Q_{a}\right|=\left|\Delta Q\right|$

$\text{relative error}=\frac{\text{true error}}{\text{true value}}=\left|\frac{\Delta Q}{Q_{t}}\right|$ $\text{percentage relative error}=\left|\frac{\Delta Q}{Q_{t}}\right|\times100$ If $$\Delta Q$$ is approximated by $$dQ$$, then the relative error is approximated by $$|dQ/Q|$$.

Example 6

The radius of a spherical tank is measured with percentage error within $$\pm2\%$$. Estimate the percentage error in the calculated volume of the tank using the measured value of the radius.

Solution 6

Let $$x$$ and $$V$$ denote the radius and the volume of the sphere, respectively. We know $V=\frac{4}{3}\pi x^{3}.$ If the error in measuring $$x$$ is denoted by $$dx=\Delta x$$, then the corresponding percentage error in the calculated volume is $$\Delta V/V$$. We may approximate $$\Delta V$$ by $$dV$$ and compute $$dV/V$$ instead of $$\Delta V/V$$: $V=\frac{4}{3}\pi x^{3}\Rightarrow dV=4\pi x^{2}\ dx$ $\frac{dV}{V}=\frac{4\pi x^{2}\ dx}{\frac{4}{3}\pi x^{3}}=3\frac{dx}{x}.$ Since the percentage error in measuring $$x$$ is $$\pm2\%$$, which means $-0.02\leq\frac{dx}{x}\leq0.02,$ the percentage error in the calculated volume is obtained by multiplying these inequalities by 3 $-0.06\leq\frac{dV}{V}=3\frac{dx}{x}\leq0.06.$ So the percentage error in the calculated volume is estimated to be within $$\pm6\%$$.

### Properties of Differentials

Since the differential of a function is the derivative of the function multiplied by the differential of the independent variable, we may multiply both sides of the familiar formulas for finding derivatives by $$dx$$ (or the differential of the independent variable) and obtain the corresponding formulas for finding differentials.

For example, if $$u=f(x)$$ and $$v=g(x)$$, the equivalent of the Sum Rule $\frac{d}{dx}(u+v)=\frac{du}{dx}+\frac{dv}{dx},$ in the notation of differentials is $d(u+v)=du+dv.$

Derivative Formula Differential Formula
$$\dfrac{d}{dx}c=0\quad$$ ($$c$$ is a constant) $$dc=0$$
$$\dfrac{d}{dx}(cu)=c\dfrac{du}{dx}$$ $$d(cu)=cdu$$
$$\dfrac{d}{dx}(u+v)=\dfrac{du}{dx}+\dfrac{dv}{dx}$$ $$d(u+v)=du+dv$$
$$\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$$ $$d(uv)=udv+vdu$$
$$\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^{2}}$$ $$d\left(\dfrac{u}{v}\right)=\dfrac{vdu-udv}{v^{2}}$$
$$\dfrac{d}{dx}(x^{r})=rx^{r-1}\quad(r\in\mathbb{R})$$ $$d(x^{r})=rx^{r-1}dx$$
$$\dfrac{d}{dx}\sin u=\cos u\dfrac{du}{dx}$$ $$d(\sin u)=\cos u\ du$$
$$\dfrac{d}{dx}\tan u=\sec^{2}u\dfrac{du}{dx}$$ $$d(\tan u)=\sec^{2}u\ du$$
$$\dfrac{d}{du}\arcsin u=\dfrac{1}{\sqrt{1-u^{2}}}$$ $$d(\arcsin u)=\dfrac{du}{\sqrt{1-u^{2}}}$$

### The Invariance of the Form of the Differential

#### Read the explanation

Let $$y=f(x)$$. Assume that $$x$$ is not an independent variable, but $$x$$ itself is a function of $$t$$. Let $$x=g(t)$$. Then $y=f(g(t))=h(t).$ If $$f,g,$$ and $$h$$ are differentiable functions, then $$dt$$ is a new independent variable with the range $$(-\infty<dt<+\infty)$$, and $$dx$$ and $$dy$$ in terms of $$t$$ and $$dt$$ are \begin{aligned} x=g(t)\quad & \Rightarrow\quad dx=g'(t)dt\\ y=h(t)\quad & \Rightarrow\quad dy=h'(t)dt\end{aligned} It follows from the Chain Rule that $\frac{dy}{dt}=h'(t)=f'(g(t))g'(t).$ Therefore, \begin{aligned} dy & =h'(t)dt\\ & =f'(\underbrace{g(t)}_{x})\underbrace{g'(t)dt}_{dx}\\ & =f'(x)dx.\end{aligned} This shows that if $$y=f(x)$$, then $dy=f'(x)dx,$ whether or not $$x$$ is the independent variable.

For example, if $$y=\sin\sqrt{x}$$, there are two ways to find $$dy$$.
Method 1: Suppose $$x$$ is the independent variable. Then \begin{aligned} dy & =\left(\dfrac{d}{dx}\sin\sqrt{x}\right)dx\\ & \begin{equation*}=(\cos\sqrt{x})\dfrac{d\sqrt{x}}{dx}dx\tag{by Chain Rule}\end{equation*} \\ & =(\cos\sqrt{x})\frac{1}{2\sqrt{x}}dx.\end{aligned} Method 2: Suppose $$u$$ is the independent variable where $$u=\sqrt{x}$$. Then \begin{aligned} dy & =\cos u\ du\\ & =\cos\sqrt{x}\ d(\sqrt{x})\quad{u=\sqrt{x},}\end{aligned} but $du=d(\sqrt{x})=\frac{1}{2\sqrt{x}}dx.$ Therefore, $dy=\cos\sqrt{x}\underbrace{\left(\frac{1}{2\sqrt{x}}dx\right)}_{d(\sqrt{x})},$ and the results of both methods are the same.

### The Chain Rule and The Derivative Rule for Inverses Are NOT Algebraic Identities

#### Read the explanation

• Although in this section we have shown that the derivative of $$y=f(x)$$ is $$dy$$ divided by $$dx$$, if $$z=g(y)$$ and $$y=f(x)$$, to justify the Chain Rule $\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}$ we cannot cancel $$dy$$’s. In other words, we can NOT say that the Chain Rule is an algebraic identity for several reasons. Firstly, we need to show that $$dz/dx$$ exists (i.e. $$g\circ f$$ is differentiable), otherwise $$dz/dx$$ would be meaningless. Secondly, we have to apply the Chain Rule to show that $$dz=g'(y)dy$$ is true even when $$y$$ is not the independent variable. Lastly, it might happen that $$dy=0$$, and in this case, we are not allowed to cancel $$dy$$ from the numerator and the denominator.

• Similarly, the Derivative Rule for Inverses which can be written as $\frac{dx}{dy}=\frac{1}{\dfrac{dy}{dx}},$ is not an algebraic identity because we first need to show that $$dx/dy$$ exists (i.e. we need to show that the inverse function is differentiable).