Linearization and Differentials (by Example)

In this article, I discuss what differentials are and how to use them. I also explain what linearization is and demonstrate how to linearize a function at a point. So, you will understand the so-called tangent line approximation with these examples.

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

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 $x=0?$

David A. Smith at Dave4Math

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.

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