We discuss linearization (or tangent line approximation) and differentials. We consider how these ideas arise naturally from the definition of the derivative and we also illustrate their use through several examples.

## Differentials

In this section differentials are motivated, defined, and then used to approximate real numbers. The linearization of a function around a point (via the tangent line) is illustrated.

Let $y=f(x)$ where $f$ is a differentiable function. The ** differential** $dy$ represents the amount that the tangent line rises or falls, whereas $\Delta y$ represents the amount that the curve $y=f(x)$ rises or falls when $x$ changes by an amount $d x.$ Since \begin{equation} \frac{d y}{d x}=\lim _{\Delta x\to 0}\frac{\Delta y}{\Delta x} \end{equation} we have \begin{equation} \frac{\Delta y}{\Delta x}\approx \frac{d y}{d x} \qquad \text{whenever} \qquad \Delta x \approx 0 \end{equation} If we take $d x=\Delta x,$ then we have $\Delta y\approx d y$ which says that the actual change in $y$ is approximately equal to the differential $d y.$

If $f\left(x_1\right)$ is a known number and it is desired to calculate an approximate value for $f\left(x_1+\Delta x\right)$ where $\Delta x$ is small, then \begin{equation} f\left(x_1+\Delta x\right)\approx f\left(x_1\right)+d y. \end{equation}

**Definition**. If $y=f(x)$ is a differentiable function then the ** differential** $dy$ is defined by the equation $d y=f'(x)d x$ where $d x$ is an independent variable.

**Example**. Find the differential for $y=\sqrt[4]{x}.$

**Solution**. For $y=\sqrt[4]{x}$ the derivative is $\frac{d y}{d x}=\frac{1}{4 x^{3/4}}$ and so the differential of $y$ is \begin{equation} dy=\left(\frac{1}{4 x^{3/4}}\right)d x \end{equation} as desired.

**Example**. Find the differential for $y=\left(x^2-2x-3\right)^{10}.$

**Solution**. For $y=\left(x^2-2x-3\right)^{10}$ the derivative is \begin{equation}\frac{d y}{d x}=10 (-2+2 x) \left(-3-2 x+x^2\right)^9 \end{equation} and so the differential of $y$ is \begin{equation} d y=\left(10 (-2+2 x) \left(-3-2 x+x^2\right)^9\right)d x. \end{equation} as desired.

**Example**. Find the differential for $y=\sqrt{x+\sqrt{2x-1}}.$

**Solution**. For $y=\sqrt{x+\sqrt{2x-1}}$ the derivative is \begin{equation} \frac{d y}{d x}=\frac{1+\sqrt{-1+2 x}}{2 \sqrt{-1+2 x} \sqrt{x+\sqrt{-1+2 x}}} \end{equation} and so the differential of $y$ is \begin{equation} d y=\left(\frac{1+\sqrt{-1+2 x}}{2 \sqrt{-1+2 x} \sqrt{x+\sqrt{-1+2 x}}}\right)d x \end{equation} as desired.

## Approximating Decimals

**Example**. Use differentials to approximate the real number $\sqrt[3]{218}.$

**Solution**. If $y=f(x)=\sqrt[3]{x},$ then $d y=\left(\frac{1}{3 x^{2/3}}\right)d x$ and using $d x=\Delta x$ $=\text{218-216}$ $=2$ the linear approximation is, \begin{align} \sqrt[3]{218} & =f(216+2) \\ & \approx f(216)+d y =6+\left(\frac{1}{3 (216)^{2/3}}\right)(2) =\frac{325}{54}=6.01852. \end{align} as desired.

**Example**. Use differentials to approximate the real number $$ \sqrt[3]{1.02}+\sqrt[4]{1.02}. $$

**Solution**. If $y=f(x)=x^{1/3}+x^{1/4},$ then $d y=\left(\frac{1}{4 x^{3/4}}+\frac{1}{3 x^{2/3}}\right) d x$ and using $d x=\Delta x$ $=\text{1.02-1}$ $=0.02$ the linear approximation is, \begin{align} \sqrt[3]{1.02}+\sqrt[4]{1.02} & =f(1+0.02) \\ & \approx f(1)+d y =2+\left(\frac{1}{4}+\frac{1}{3}\right)(0.02) =\frac{1207}{600}=2.011\overline{6} \end{align}

as desired.

## Linearization

**Definition**. The approximation \begin{equation} f(x)\approx f(a)+f'(a)(x-a)\end{equation} is called the ** linear approximation** (or sometimes the

**) of $f$ at $a$ and the function \begin{equation} L(x)=f(a)+f'(a)(x-a) \end{equation} is called the**

*tangent line approximation***of $f$ at $a.$**

*linearization*The equation of the tangent line to the curve $y=f(x)$ at $(a, f(a))$ is $$ y=f(a)+f'(a)(x-a) $$ which is $$ y=f(a)+f'(a)d x $$ so that in fact we have $y=f(a)+d y.$ Thus when using differentials to approximate, that is, when using $$f(x+\Delta x)\approx f(x)+d y $$ to approximate we are using the tangent line at $(a, f(a))$ as an approximation to the curve $y=f(x)$ when $a$ is near $x.$

Next we find the linearization for a given function at a given value.

**Example**. Find the linearization of $\displaystyle f(x)=\frac{1}{\sqrt{2+x}}$ at $x_1=0.$

**Solution**. The linearization of the function $f$ at $0$ is \begin{align} L(x) & = f\left(x_1\right) + f’\left(x_1\right)\left(x-x_1\right) \\ L(x) & =f(0)+\left(-\frac{1}{2 (2+0)^{3/2}}\right) \\ L(x) & =\frac{1}{\sqrt{2}}+\left(-\frac{1}{4 \sqrt{2}}\right)x \end{align} Therefore, we have the linear approximation \begin{equation} \frac{1}{\sqrt{2+x}}\approx \frac{\sqrt{2}}{2}+\left(-\frac{\sqrt{2}}{8}\right)x \end{equation} when $x$ is near 0.

**Example**. Find the linearization of $\displaystyle f(x)=\frac{1}{(1+2x)^4}$ at $x_1=0.$

**Solution**. The linearization of the function $f$ at $0$ is \begin{align} L(x) & = f\left(x_1\right) + f’\left(x_1\right)\left(x-x_1\right) \\ & = f(0)+\left(-\frac{8}{(1+2 (0))^5} \right) \\ & =1-8x. \end{align} Therefore, we have the linear approximation

\begin{equation} \frac{1}{(1+2x)^4}\approx 1-8x \end{equation} when $x$ is near 0.

## Exercises on Linearization and Differentials

**Exercise**. Given $f(x)=x^2+2x,$ $x_0=1,$ and $\Delta x=0.1,$ find the change $\Delta f=f\left(x_0+\Delta x\right)-f\left(x_0\right)$ and the value of the estimate $df=f’\left(x_0\right)dx.$

**Exercise**. Given $f(x)=x^4,$ $x_0=1,$ and $\Delta x=0.1,$ find the change $\Delta f=f\left(x_0+\Delta x\right)-f\left(x_0\right)$ and the value of the estimate $df=f’\left(x_0\right)dx.$

**Exercise**. Given $f(x)=x^3-2x+3$ $x_0=2,$ and $\Delta x=0.1,$ find the change $\Delta f=f\left(x_0+\Delta x\right)-f\left(x_0\right)$ and the value of the estimate $$ df=f’\left(x_0\right)dx. $$

**Exercise**. Write a differential formula that estimates the change in the volume $V=\frac{4}{3}\pi r^3$ of a sphere when the radius changes from $r_0$ to $r_0+dr.$

**Exercise**. Write a differential formula that estimates the change in the lateral surface area $S=\pi r\sqrt{r^2+h^2}$ of a right circular cone when the radius changes from $r_0$ to $r_0+dr.$

**Exercise**. Find the differential $dy$ of the following functions.

$(1) \quad \displaystyle y=f(x)=\frac{1}{3}\cos \left(\frac{6\pi x-1}{2}\right)$

$(2) \quad \displaystyle y=f(x)=\sqrt{\sqrt{x}+\frac{1}{\sqrt{x}}}.$

**Exercise**. Find the differential $dy$ of each of the following functions.

$(1) \quad \displaystyle y=f(x)=\frac{2x}{1+x^2}$

$(2) \quad \displaystyle x y^2-4x^{3/2}-y=0$

$(3) \quad \displaystyle y=f(x)=4 \tan \left(\frac{x^2}{3}\right)$

$(4) \quad \displaystyle y=f(x)=2 \cot \left(\frac{1}{\sqrt{x}}\right)$

$(5) \quad \displaystyle y=f(x)=x e^{-x}$

$(6) \quad \displaystyle y=f(x)=\ln \left(1+x^2\right)$

**Exercise**. Use differentials to find an approximate value for the following real numbers

$(1) \quad \cos 31.5{}^{\circ}$

$(2) \quad \sqrt[4]{624}$

$(3) \quad (2.99)^3.$

**Exercise**. Find the linearization for each of the following functions at the given value of $x.$

$(1) \quad \displaystyle f(x)=\frac{1}{\sqrt{2+x}}$ at $x_1=0$

$(2) \quad \displaystyle f(x)=\frac{1}{(1+2x)^4}$ at $x_1=0$

$(3) \quad \displaystyle f(x)=x+\frac{1}{x}$ at $x=a$

$(4) \quad \displaystyle f(x)=x^2$ at $x=a$

$(5) \quad \displaystyle f(x)=\cos x$ at $x=a$

$(6) \quad \displaystyle f(x)=\tan x$ at $x=a$

$(7) \quad \displaystyle f(x)=e^x$ at $x=a$

$(8) \quad \displaystyle f(x)=\ln (1+x)$ at $x=a$

$(9) \quad \displaystyle f(x)=x+\frac{1}{x}$ at $x=a$

$(10) \quad \displaystyle f(x)=\frac{x}{x+1}$ at $x_0=1.3$

**Exercise**. Show that the linearization of $f(x)=(1+x)^k$ at $x=0$ is $L(x)=1+ k x$ for a constant $k.$

**Exercise**. Find the linearization of $f(x)=\sqrt{x+1}+\sin x$ at $x=0.$ How is it related to the individual linearization of $\sqrt{x+1}$ and $\sin

x$ at $x=0?$