# Calculus: Derivatives IV

## Linear Approximation

The derivative of a function provides the slope of the line that is tangent to the function at a given point. Given the slope and the coordinates of the point, it is possible to construct the tangent line itself. This tangent line happens to be a useful *approximation* of the function itself, at least for points close to the intersection. Making use of a **linear approximation** like this is convenient for when calculating the value of the original function is messy or difficult at nearby points. For example, computer algorithms often use linear approximations when calculating the exact answer would take too long.

As a conceptual aid, consider the fact that the earth is round. Yet, if you "zoom in" close enough to human scale, you will see that the earth looks remarkably flat. That is to say, the earth looks very *linear*. In fact, assuming the earth is flat is useful for all kinds of activities that do not take place on a scale large enough for the curvature of the earth to make much of a difference. So it is with mathematical curves - if you zoom in enough, curves look more and more linear, and linear approximations become more and more accurate.

Given a point on a function $(x_0,f(x_0))$, there are two similar techniques that can be used to approximate the value of nearby a nearby point, $(x_1,f(x_1))$. The first involves constructing the tangent line itself and using the value of the tangent line at the nearby point as the approximation. Here are the following steps for carrying out this first technique:

- Take the derivative of the original function $f(x)$.
- Calculate the value of the derivative $f'(x)$ at $x_0$.
- Using $x_0, f(x_0)$, and $f'(x_0)$, construct the equation of the tangent line, call it $g(x)$.
- Approximate the nearby value of $f(x_1)$ by calculating $g(x_1)$.

The second technique is similar to the first, but it involves using the derivative to "jump" directly to the nearby point without having to calculate the rest of the tangent line equation:

- Take the derivative of the original function $f(x)$.
- Calculate the difference between the $x_0$ and $x_1$, and call it $\Delta x$. That is to say, let $\Delta x = x_1 - x_2$.
- Remember that the slope of a line is $\dfrac{\text{rise}}{\text{run}}$, or in our case, $f'(x_0)=\dfrac{\Delta y}{\Delta x}$. Since we've calculated the run, $\Delta x$, we simply multiply $f'(x_0)$ by $\Delta x$ to get $\Delta y$.
- Approximate $f(x_1)$ as $f(x_1) = f(x_0) + f'(x_0) \cdot \Delta x$.

## Problems

Given the fact that $\ln(1)=0$, Approximate the value of $\ln(1.1)$

**Step 1:**Take the derivative

$\dfrac{d}{dx}\ln(x) = \dfrac{1}{x}$**Step 2:**Calculate $\Delta x$

$\Delta x = 1.1 - 1 \\ \Delta x = 0.1$**Step 3:**Approximate using the tangent line:

$\ln(1.1) \approx \ln(1) + \left. \dfrac{d}{dx}(\ln(x))\right|_{x=1} \Delta x \\ \ln(1.1) \approx 0 + \dfrac{1}{1} 0.1 \\ \ln(1.1) \approx 0.1 \\ $

The bar in step 3 means to evaluate the expression to the left where $x=1$. The whole term means "Take the derivative with respect to $x$ of $\ln(x)$ and then evaluate it at $x=1$."Let $f(x)=x^2 + 1$. Given that $f(2)=5$, calculate $f(2.05)$.

**Step 1:**Calculate the derivative at the starting point

$f'(x) = \dfrac{d}{dx}(x^2+1) \\ f'(x) = 2x + 1 \\ f'(2) = 2(2) + 1 f'(2) = 5 $**Step 2:**Calculate $\Delta x$

$\Delta x = 2.05 - 2 \\ \Delta x = 0.05$**Step 3:**Approximate using the tangent line:

$f(2.05) \approx f(2) + f'(2) \Delta x \\ f(2.05) \approx 5 + 5 \cdot 0.05 \\ f(2.05) \approx 5.25 \\ $- Let $f(x)=x^3 - 4x$. Given that $f(-1)=3$, calculate $f(-1.1)$.
**Step 1:**Calculate the derivative at the starting point

$f'(x) = \dfrac{d}{dx}(x^3-4x) \\ f'(x) = 3x^2 - 4 \\ f'(2) = 3(-1)^2 - 4 f'(2) = -1 $**Step 2:**Calculate $\Delta x$

$\Delta x = -1.1 - (-1) \\ \Delta x = -0.1$**Step 3:**Approximate using the tangent line:

$f(2.05) \approx f(-1) + f'(-1) \Delta x \\ f(2.05) \approx 3 + (-1) \cdot (-0.1) \\ f(2.05) \approx 3.1 \\ $ Let $f(x)=x^2 - 3x + 1$. Given that $f(0)=1$, approximate $f(0.1)$.

**Step 1:**Calculate the derivative at the starting point:

$f'(x) = \dfrac{d}{dx}\left(x^2 - 3x + 1\right) \\ f'(x) = 2x - 3 \\ f'(0) = 2(0) - 3 \\ f'(0) = -3 \\ $**Step 2:**Calculate $\Delta x$:

$\Delta x = 0.1 - 0 \\ \Delta x = 0.1$**Step 3:**Approximate using the tangent line:

$f(0.1) \approx f(0) + f'(0)\Delta x \\ f(0.1) \approx 1 + (-3)(0.1) \\ f(0.1) \approx 1 - 0.3 \\ f(0.1) \approx 0.7 \\ $Let $f(x)=\sin(10x)$. Given that $f(0)=0$, calculate $f(0.1)$.

**Step 1:**Calculate the derivative at the starting point

$f'(x) = \dfrac{d}{dx}\sin(10x) \\ f'(x) = 10\cos(10x) \\ f'(0) = 10\cos(10\cdot 0) \\ f'(0) = 10\cdot 1 \\ f'(0) = 10 \\ $**Step 2:**Calculate $\Delta x$

$\Delta x = 0.1 - 0 \\ \Delta x = 0.1$**Step 3:**Approximate using the tangent line:

$f(0.1) \approx f(0) + f'(0) \Delta x \\ f(0.1) \approx 0 + 10 \cdot 0.1 \\ f(0.1) \approx 1 \\ $Let $f(x)=x^4 - \dfrac{1}{2}x$. Given that $f(2)=15$, calculate $f(2.2)$.

**Step 1:**Calculate the derivative at the starting point

$f'(x) = \dfrac{d}{dx}\left(x^4 - \dfrac{1}{2}x\right) \\ f'(x) = 4x^3 - \dfrac{1}{2} \\ f'(2) = 4(2)^3 - \dfrac{1}{2} \\ f'(2) = 32 - \dfrac{1}{2} \\ f'(2) = 31.5 \\ $**Step 2:**Calculate $\Delta x$

$\Delta x = 2.2 - 2 \\ \Delta x = 0.2$**Step 3:**Approximate using the tangent line:

$f(2.2) \approx f(2) + f'(2) \Delta x \\ f(2.2) \approx 15 + 31.5 \cdot 0.2 \\ f(2.2) \approx 21.3 \\ $Let $f(x) = \tan(x)$. Given that $f\left(\dfrac{\pi}{4}\right) = 1$, approximate $f(1)$.

**Step 1:**Calculate the derivative at the starting point:

$f'(x) = \dfrac{d}{dx}\tan(x) \\ f'(x) = \sec^2(x) \\ f'\left(\dfrac{\pi}{4}\right) = \sec^2\left(\dfrac{\pi}{4}\right) \\ f'\left(\dfrac{\pi}{4}\right) = 2 \\ $**Step 2:**Calculate $\Delta x$:

$\Delta x = 1 - \dfrac{\pi}{4}$**Step 3:**Approximate using the tangent line:

$f(1) \approx f\left(\dfrac{\pi}{4}\right) + f'\left(\dfrac{\pi}{4}\right)\Delta x \\ f(1) \approx 1 + 2\left(1 - \dfrac{\pi}{4}\right) \\ f(1) \approx 1.4292 \\ $Let $f(x)=\dfrac{1}{x+1}$. Given that $f(3)=\dfrac{1}{4}$, approximate $f(3.1)$.

**Step 1:**Calculate the derivative at the starting point:

$f'(x) = \dfrac{d}{dx}\dfrac{1}{x+1} \\ f'(x) = \dfrac{-1}{(x+1)^2} \\ f'(3) = \dfrac{-1}{(3+1)^2} \\ f'(3) = -\dfrac{1}{16} \\ $**Step 2:**Calculate $\Delta x$:

$\Delta x = 3.1 - 3 \\ \Delta x = 0.1$**Step 3:**Approximate using the tangent line:

$f(3.1) \approx f(3) + f'(3)\Delta x \\ f(3.1) \approx \dfrac{1}{4} - \left(\dfrac{1}{16}\right)(0.1) \\ f(3.1) \approx \dfrac{1}{4} - \dfrac{1}{160} \\ f(3.1) \approx \dfrac{39}{160} \\ f(3.1) \approx 0.24375 \\ $Let $f(x)=\dfrac{x}{x+1}$. Given that $f(1)=\dfrac{1}{2}$, approximate $f(1.1)$.

**Step 1:**Calculate the derivative at the starting point:

$f'(x) = \dfrac{d}{dx}\dfrac{x}{x+1} \\ f'(x) = \dfrac{(x+1)(1)-x(1)}{(x+1)^2} \\ f'(x) = \dfrac{1}{(x+1)^2} \\ f'(1) = \dfrac{1}{(1+1)^2} \\ f'(1) = \dfrac{1}{2} \\ $**Step 2:**Calculate $\Delta x$:

$\Delta x = 1.1 - 1 \\ \Delta x = 0.1$**Step 3:**Approximate using the tangent line:

$f(1.1) \approx f(1) + f'(3)\Delta x \\ f(1.1) \approx \dfrac{1}{2} + \left(\dfrac{1}{2}\right)(0.1) \\ f(1.1) \approx \dfrac{1}{2} + \dfrac{1}{20} \\ f(1.1) \approx \dfrac{11}{20} \\ f(1.1) \approx 0.55 \\ $Let $f(x)=\dfrac{1}{x^2}$. Given that $f(-1)=1$, approximate $f(-1.1)$.

**Step 1:**Calculate the derivative at the starting point:

$f'(x) = \dfrac{d}{dx}\dfrac{1}{x^2} \\ f'(x) = \dfrac{-2}{x^3} \\ f'(-1) = \dfrac{-2}{(-1)^3} \\ f'(-1) = 2 \\ $**Step 2:**Calculate $\Delta x$:

$\Delta x = -1.1 - (-1) \\ \Delta x = -0.1$**Step 3:**Approximate using the tangent line:

$f(-1.1) \approx f(-1) + f'(-1)\Delta x \\ f(-1.1) \approx 1 + (2)(-0.1) \\ f(-1.1) \approx 1 - 0.2 \\ f(-1.1) \approx 0.8 \\ $