Derivatives - L Hopital Rule Simple Definition

Derivatives - L'Hôpital's Rule Simple Definition

</section>
<section style="font-size: 50px";>
<h2 id="derivatives---lhôpitals-rule-simple-definition">Derivatives - L’Hôpital’s Rule Simple Definition</h2>
<pre><code>            * L'Hôpital's Rule is a technique used to evaluate limits of indeterminate forms of the type 0/0 or ∞/∞.
            
            * It helps in finding the limit of a function as it approaches a certain value or as it approaches infinity or negative infinity.
            
            * L'Hôpital's Rule is based on differentiation, and it allows us to resolve such indeterminate forms by taking the derivatives of the numerator and the denominator separately.
            
            * Let's understand this rule with some examples.
            
            * L'Hôpital's Rule is named after the French mathematician Guillaume de l'Hôpital.
        </textarea>
    </section>
    </textarea> </section> <section data-markdown data-separator='' >
        <textarea data-template>
</code></pre>
</section>
<section style="font-size: 50px";>
<h2 id="example-1-limit-of-x2--3x--4x--2-as-x-approaches-2">Example 1: Limit of (x^2 + 3x) / (4x + 2) as x approaches 2</h2>
<pre><code>            * Let's apply L'Hôpital's Rule to find the limit of the given function.
            
            * First, we need to differentiate both the numerator and the denominator.
            
            * Differentiating the numerator: `d/dx (x^2 + 3x) = 2x + 3`
            
            * Differentiating the denominator: `d/dx (4x + 2) = 4`
            
            * Now, our new function is (2x + 3) / 4
            
            * Taking the limit as x approaches 2, we substitute x = 2 in the new function.
            
            * The limit evaluates to (2(2) + 3) / 4 = (4 + 3) / 4 = 7/4.
        </textarea>
    </section>
    </textarea> </section> <section data-markdown data-separator='' >
        <textarea data-template>
</code></pre>
</section>
<section style="font-size: 50px";>
<h2 id="example-2-limit-of-ex---1--x-as-x-approaches-0">Example 2: Limit of (e^x - 1) / x as x approaches 0</h2>
<pre><code>            * Let's use L'Hôpital's Rule to find the limit of the given function.
            
            * We differentiate the numerator and the denominator separately.
            
            * Differentiating the numerator: `d/dx (e^x - 1) = e^x`
            
            * Differentiating the denominator: `d/dx (x) = 1`
            
            * The new function becomes e^x / 1 = e^x
            
            * Taking the limit as x approaches 0, we substitute x = 0 in the new function.
            
            * The limit evaluates to e^0 = 1.
        </textarea>
    </section>
    </textarea> </section> <section data-markdown data-separator='' >
        <textarea data-template>
</code></pre>
</section>
<section style="font-size: 50px";>
<h2 id="example-3-limit-of-sinx--x-as-x-approaches-0">Example 3: Limit of sin(x) / x as x approaches 0</h2>
<pre><code>            * By using L'Hôpital's Rule, we can find the limit of the given function.
            
            * Differentiating the numerator: `d/dx (sin(x)) = cos(x)`
            
            * Differentiating the denominator: `d/dx (x) = 1`
            
            * The new function becomes cos(x) / 1 = cos(x)
            
            * Taking the limit as x approaches 0, we substitute x = 0 in the new function.
            
            * The limit evaluates to cos(0) = 1.
        </textarea>
    </section>
    </textarea> </section> <section data-markdown data-separator=''  >
        <textarea data-template>
</code></pre>
</section>
<section style="font-size: 50px";>
<h2 id="example-4-limit-of-1---cosx--x2-as-x-approaches-0">Example 4: Limit of (1 - cos(x)) / x^2 as x approaches 0</h2>
<pre><code>            * Let's apply L'Hôpital's Rule to evaluate the limit of the given function.
            
            * Differentiating the numerator: `d/dx (1 - cos(x)) = sin(x)`
            
            * Differentiating the denominator: `d/dx (x^2) = 2x`
            
            * The new function becomes sin(x) / 2x
            
            * Taking the limit as x approaches 0, we substitute x = 0 in the new function.
            
            * The limit evaluates to sin(0) / 2(0) = 0 / 0
            
            * Since the new function is still indeterminate, we can apply L'Hôpital's Rule again by differentiating the numerator and the denominator.
        </textarea>
    </section>
    </textarea> </section> <section data-markdown data-separator=''  >
        <textarea data-template>
</code></pre>
</section>
<section style="font-size: 50px";>
<h2 id="example-4-continued">Example 4: (Continued)</h2>
<pre><code>            * Differentiating the numerator: `d/dx (sin(x)) = cos(x)`
            
            * Differentiating the denominator: `d/dx (2x) = 2`
            
            * The new function becomes cos(x) / 2
            
            * Taking the limit as x approaches 0, we substitute x = 0 in the new function.
            
            * The limit evaluates to cos(0) / 2 = 1 / 2.
            
            * Therefore, the limit of the given function as x approaches 0 is 1/2.
        </textarea>
    </section>
     </textarea> </section> <section data-markdown data-separator=''  >
        <textarea data-template>
</code></pre>
</section>
<section style="font-size: 50px";>
<h2 id="summary">Summary</h2>
<pre><code>            * L'Hôpital's Rule is an essential tool for evaluating limits of indeterminate forms.
            
            * It allows us to resolve indeterminate forms of the type 0/0 or ∞/∞ by taking the derivatives of the numerator and the denominator separately.
            
            * The rule is named after the French mathematician Guillaume de l'Hôpital.
            
            * We can apply L'Hôpital's Rule multiple times if necessary.
            
            * It is crucial to understand the concept and apply the rule correctly while solving problems related to limits.
        </textarea>
    </section>
</body>
</code></pre>
</html>
</section>
<section style="font-size: 50px";>
<ol start="11">
<li>Derivative Rules</li>
</ol>
<ul>
<li>The power rule: $\frac{d}{dx} (x^n) = nx^{n-1}$</li>
<li>The constant rule: $\frac{d}{dx} (c) = 0$</li>
<li>The sum rule: $\frac{d}{dx} (f(x) + g(x)) = \frac{d}{dx} (f(x)) + \frac{d}{dx} (g(x))$</li>
<li>The product rule: $\frac{d}{dx} (f(x) \cdot g(x)) = f’(x) \cdot g(x) + f(x) \cdot g’(x)$</li>
<li>The quotient rule: $\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f’(x) \cdot g(x) - f(x) \cdot g’(x)}{(g(x))^2}$</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="12">
<li>Chain Rule</li>
</ol>
<ul>
<li>The chain rule allows us to find the derivative of composite functions.</li>
<li>If $y = f(g(x))$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$</li>
<li>Example: Find $\frac{dy}{dx}$ if $y = (2x^3 + 5x + 1)^4$</li>
<li>Solution: Let $u = 2x^3 + 5x + 1$. Applying the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 4u^3 \cdot (6x^2 + 5)$</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="13">
<li>Implicit Differentiation</li>
</ol>
<ul>
<li>Implicit differentiation is used to find the derivative of a function expressed implicitly.</li>
<li>To differentiate implicitly, differentiate each term with respect to $x$ and treat $y$ as a function of $x$.</li>
<li>Example: Find $\frac{dy}{dx}$ if $x^2 + y^2 = 25$</li>
<li>Solution: Differentiating both sides with respect to $x$, we get $2x + 2y \cdot \frac{dy}{dx} = 0$. Solving for $\frac{dy}{dx}$, we have $\frac{dy}{dx} = \frac{-2x}{2y} = \frac{-x}{y}$.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="14">
<li>Related Rates</li>
</ol>
<ul>
<li>Related rates problems involve finding how fast one quantity is changing with respect to another related quantity.</li>
<li>To solve related rates problems, identify the relevant variables, differentiate with respect to time, and apply appropriate equations.</li>
<li>Example: A ladder is leaning against a wall. The foot of the ladder is sliding away from the wall at a rate of 2 ft/s. If the top of the ladder is sliding down the wall at a rate of 3 ft/s, how fast is the length of the ladder changing when the foot of the ladder is 12 ft away from the wall?</li>
<li>Solution: Let $x$ be the distance of the foot of the ladder from the wall, and let $y$ be the height of the ladder on the wall. We are given $\frac{dx}{dt} = 2$ ft/s and $\frac{dy}{dt} = -3$ ft/s. We need to find $\frac{dl}{dt}$ when $x = 12$ ft. Using the Pythagorean theorem, $x^2 + y^2 = l^2$, where $l$ is the length of the ladder. Differentiating both sides with respect to $t$, we get $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2l\frac{dl}{dt}$. Substituting the known values and solving for $\frac{dl}{dt}$, we find that $\frac{dl}{dt} = -\frac{8}{5}$ ft/s.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="15">
<li>Optimization</li>
</ol>
<ul>
<li>Optimization problems involve finding the maximum or minimum value of a quantity subject to certain constraints.</li>
<li>To solve optimization problems, identify the objective function and the constraints, find the derivative of the objective function, set it equal to zero to find critical points, and check the endpoints and critical points for the maximum or minimum value.</li>
<li>Example: A rectangular plot of land is to be fenced on three sides, with the fourth side along a river. If there is 60 meters of fencing available, what dimensions will maximize the area of the plot?</li>
<li>Solution: Let $x$ represent the length of the plot and $y$ represent the width. The objective function is $A = xy$, and the constraint is $2x + y = 60$. Solving for $y$, we get $y = 60 - 2x$. Substituting this into the objective function, we have $A = x(60 - 2x)$. Differentiating $A$ with respect to $x$, setting it equal to zero, and solving for $x$, we find $x = 15$ and $x = 30$. Checking the endpoints and critical points, we find that $x = 15$ gives the maximum area. Therefore, the dimensions that maximize the area of the plot are 15 m by 30 m.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="16">
<li>Derivative of Exponential Functions</li>
</ol>
<ul>
<li>The derivative of the exponential function $f(x) = e^x$ is equal to itself: $\frac{d}{dx}(e^x) = e^x$.</li>
<li>Example: Find the derivative of $f(x) = 3e^x$.</li>
<li>Solution: Using the constant rule, $\frac{d}{dx}(3e^x) = 3 \cdot \frac{d}{dx}(e^x) = 3 \cdot e^x = 3e^x$.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="17">
<li>Derivative of Logarithmic Functions</li>
</ol>
<ul>
<li>The derivative of the natural logarithm function $f(x) = \ln(x)$ is $\frac{1}{x}$: $\frac{d}{dx}(\ln(x)) = \frac{1}{x}$.</li>
<li>Example: Find the derivative of $g(x) = \ln(5x)$.</li>
<li>Solution: Using the chain rule, we have $\frac{d}{dx}(\ln(5x)) = \frac{1}{5x} \cdot \frac{d}{dx}(5x) = \frac{1}{5x} \cdot 5 = \frac{1}{x}$.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="18">
<li>Derivative of Trigonometric Functions</li>
</ol>
<ul>
<li>The derivative of the sine function $f(x) = \sin(x)$ is the cosine function: $\frac{d}{dx}(\sin(x)) = \cos(x)$.</li>
<li>The derivative of the cosine function $g(x) = \cos(x)$ is the negative sine function: $\frac{d}{dx}(\cos(x)) = -\sin(x)$.</li>
<li>Example: Find the derivative of $h(x) = \sin(x) + \cos(x)$.</li>
<li>Solution: Using the sum rule, $\frac{d}{dx}(\sin(x) + \cos(x)) = \frac{d}{dx}(\sin(x)) + \frac{d}{dx}(\cos(x)) = \cos(x) - \sin(x)$.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="19">
<li>Derivative of Inverse Functions</li>
</ol>
<ul>
<li>The derivative of an inverse function $f^{-1}(x)$ is equal to the reciprocal of the derivative of the original function: $\frac{d}{dx}(f^{-1}(x)) = \frac{1}{\frac{d}{dx}(f(x))}$.</li>
<li>Example: Find the derivative of $f^{-1}(x)$ if $f(x) = \log_{10}(x)$.</li>
<li>Solution: First, find $\frac{d}{dx}(f(x)) = \frac{1}{x \ln(10)}$. Then, the derivative of $f^{-1}(x)$ is $\frac{1}{\frac{1}{x \ln(10)}} = x \ln(10)$.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="20">
<li>Mean Value Theorem</li>
</ol>
<ul>
<li>The Mean Value Theorem states that if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one point $c \in (a, b)$ such that $f’(c) = \frac{f(b) - f(a)}{b - a}$.</li>
<li>Example: Find the point(s) that satisfies the Mean Value Theorem for the function $f(x) = x^2 - 2x + 2$ on the interval $[-1, 2]$.</li>
<li>Solution: First, find $f(-1) = 6$ and $f(2) = 4$. Then, find $f’(-1) = 4$ and $f’(2) = 2$. Notice that $f’(-1) \neq f’(2)$. From the Mean Value Theorem, there exists at least one point $c \in (-1, 2)$ such that $f’(c) = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{4 - 6}{2 + 1} = -\frac{2}{3}$.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="21">
<li>Tangent and Normal Lines</li>
</ol>
<ul>
<li>The tangent line to a curve at a specific point is a line that touches the curve at that point and has the same slope as the curve at that point.</li>
<li>The normal line to a curve at a specific point is a line that is perpendicular to the tangent line at that point.</li>
<li>The equations of the tangent and normal lines can be found using the point-slope form: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="22">
<li>Implicit Differentiation and Second Derivatives</li>
</ol>
<ul>
<li>Implicit differentiation is used to find the derivative of a function expressed implicitly.</li>
<li>To differentiate implicitly, differentiate each term with respect to $x$ and treat $y$ as a function of $x$.</li>
<li>To find the second derivative implicitly, differentiate the first derivative with respect to $x$.</li>
<li>Example: Find the equation of the tangent line and the normal line to the curve $x^2 + y^2 = 1$ at the point $(1/2, \sqrt{3}/2)$.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="23">
<li>Rolle’s Theorem and the Mean Value Theorem</li>
</ol>
<ul>
<li>Rolle’s Theorem states that if a function is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one point $c \in (a, b)$ such that $f’(c) = 0$.</li>
<li>The Mean Value Theorem states that if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one point $c \in (a, b)$ such that $f’(c) = \frac{f(b) - f(a)}{b - a}$.</li>
<li>These theorems are important tools in calculus to study the behavior of functions and find critical points and intervals of increase and decrease.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="24">
<li>Concavity and Points of Inflection</li>
</ol>
<ul>
<li>The concavity of a graph determines whether the graph is facing upward or downward.</li>
<li>A graph is concave upward if its second derivative is positive, indicating that the graph is curved like a “U”.</li>
<li>A graph is concave downward if its second derivative is negative, indicating that the graph is curved like an “n”.</li>
<li>Points of inflection occur where the concavity of a graph changes from concave upward to concave downward or vice versa.</li>
<li>To find points of inflection, set the second derivative equal to zero and solve for $x$.</li>
</ul>
</section>
<section style="font-size: 50px";>
<ol start="25">
<li>Optimization and Related Rates</li>
</ol>
<ul>
<li>Optimization problems involve finding the maximum or minimum value of a quantity subject to certain constraints.</li>
<li>To solve optimization problems, identify the objective function and the constraints, find the derivative of the objective function, set it equal to zero to find</li>
</ul>
</section>

</div>
				</div>

function transformCallouts() {
      var slides = document.querySelectorAll('.reveal .slides section');
      slides.forEach(function (slide) {
        var data = document.getElementsByTagName("Success");
        for (let i = 0; i < data.length; i++) {
          data[i].classList.add("callout");
          data[i].classList.add("callout-success");
        }

var data = document.getElementsByTagName("Default");
        for (let i = 0; i < data.length; i++) {
          data[i].classList.add("callout");
          data[i].classList.add("callout-default");
        }

var data = document.getElementsByTagName("Warning");
        for (let i = 0; i < data.length; i++) {
          data[i].classList.add("callout");
          data[i].classList.add("callout-warning");
        }

var data = document.getElementsByTagName("Primary");
        for (let i = 0; i < data.length; i++) {
          data[i].classList.add("callout");
          data[i].classList.add("callout-primary");
        }

var data = document.getElementsByTagName("Danger");
        for (let i = 0; i < data.length; i++) {
          data[i].classList.add("callout");
          data[i].classList.add("callout-danger");
        }
      });
    };

function addrfit() {
      var slides = document.querySelectorAll('.reveal');
      slides.forEach(function (slide) {
        var temp = document.getElementsByTagName("h2");

if (temp.length > 0) {
          for (let i = 0; i < temp.length; i++) {
            temp[i].classList.add("r-fit-text");
          }
        }

var temp3 = document.getElementsByTagName("h1");

if (temp3.length > 0) {
          for (let i = 0; i < temp3.length; i++) {
            temp3[i].classList.add("r-fit-text");
          }
        }

var temp6 = document.getElementsByClassName("text-container");

if (temp6.length > 0) {
          for (let i = 0; i < temp6.length; i++) {
            var charcount = 0;
            var ptags = temp6[i].getElementsByTagName("p");
            try {
              for (let j = 0; j < ptags.length; j++) {
                temp = ptags[j].outerText.replaceAll(" ", "");
                temp = temp.replaceAll("\n", "");
                charcount += temp.length;
              }
            } catch (err) {
              console.log(err);
            }

var font = 35;

if (charcount > 50 && charcount < 100) {
              font = 35;
            } else if (charcount > 100 && charcount < 120) {
              font = 33;
            } else if (charcount > 120 && charcount < 150) {
              font = 33;
            } else if (charcount > 150 && charcount < 170) {
              font = 31;
            } else if (charcount > 170 && charcount < 200) {
              font = 30;
            } else if (charcount > 200 && charcount < 230) {
              font = 29;
            } else if (charcount > 230 && charcount < 250) {
              font = 28;
            } else if (charcount > 250 && charcount < 270) {
              font = 27;
            } else if (charcount > 270 && charcount < 300) {
              font = 25;
            } else if (charcount > 300 && charcount < 330) {
              font = 23;
            } else if (charcount > 330 && charcount < 360) {
              font = 24;
            } else if (charcount > 360 && charcount < 390) {
              font = 22;
            } else if (charcount > 390 && charcount < 420) {
              font = 21;
            } else if (charcount > 420) {
              font = 20;
            }

temp6[i].style.fontSize = `${font}px`;
          }
        }
      })
    }

Reveal.addEventListener('ready', addrfit);
    Reveal.addEventListener('ready', transformCallouts);
  </script>