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- https://math.libretexts.org/Bookshelves/Calculus/CLP-1_Differential_Calculus_(Feldman_Rechnitzer_and_Yeager)/03%3A_Derivatives/3.06%3A_Using_the_Arithmetic_of_Derivatives__ExamplesIn this section we illustrate the computation of derivatives using the arithmetic of derivatives — Theorems 2.4.2, 2.4.3 and 2.4.5. To make it clear which rules we are using during the examples we wil...In this section we illustrate the computation of derivatives using the arithmetic of derivatives — Theorems 2.4.2, 2.4.3 and 2.4.5. To make it clear which rules we are using during the examples we will note which theorem we are using:
- https://math.libretexts.org/Courses/Prince_Georges_Community_College/MAT_2160%3A_Applied_Calculus_I/01%3A_The_Derivative/1.07%3A_Higher_Order_DerivativesThe operation of differentiation takes as input one function, f(x), and produces as output another function, f′(x). Now f′(x) is once again a function. So we can differentiat...The operation of differentiation takes as input one function, f(x), and produces as output another function, f′(x). Now f′(x) is once again a function. So we can differentiate it again, assuming that it is differentiable, to create a third function, called the second derivative of f. And we can differentiate the second derivative again to create a fourth function, called the third derivative of f. And so on.
- https://math.libretexts.org/Bookshelves/Calculus/CLP-1_Differential_Calculus_(Feldman_Rechnitzer_and_Yeager)/06%3A_Appendix/6.01%3A_A-_High_School_Material/6.1.12%3A_A.12_PowersIn the following, x and y are arbitrary real numbers, and q is an arbitrary constant that is strictly bigger than zero. qx+y=qxqy, qx−y=qxqy \(q^{-x}=\fra...In the following, x and y are arbitrary real numbers, and q is an arbitrary constant that is strictly bigger than zero. qx+y=qxqy, qx−y=qxqy q−x=1qx lim \lim\limits_{x\rightarrow-\infty}q^x=0 if q \gt 1 \lim\limits_{x\rightarrow\infty}q^x=0\text{,} \lim\limits_{x\rightarrow-\infty}q^x=\infty if 0 \lt q \lt 1
- https://math.libretexts.org/Bookshelves/Calculus/CLP-1_Differential_Calculus_(Feldman_Rechnitzer_and_Yeager)/06%3A_Appendix/6.02%3A_B-_Origin_of_Trig_Area_and_Volume_Formulas
- https://math.libretexts.org/Bookshelves/Calculus/CLP-1_Differential_Calculus_(Feldman_Rechnitzer_and_Yeager)/01%3A_The_basics/1.05%3A_Parsing_FormulasConsider the formula
- https://math.libretexts.org/Bookshelves/Calculus/CLP-1_Differential_Calculus_(Feldman_Rechnitzer_and_Yeager)/06%3A_Appendix/6.03%3A_C-_Root_Finding/6.3.04%3A_C.4_The_secant_methodIt avoids the use of the derivative by approximating f'(x) by \frac{f(x+h)-f(x)}{h} for some h\text{.} That is, it approximates the tangent line to f at x by a secant line for \(f\...It avoids the use of the derivative by approximating f'(x) by \frac{f(x+h)-f(x)}{h} for some h\text{.} That is, it approximates the tangent line to f at x by a secant line for f that passes through x\text{.} To limit the number of evaluations of f(x) required, it uses x=x_{n-1} and x+h=x_n\text{.} Here is how it works.
- https://math.libretexts.org/Bookshelves/Calculus/CLP-1_Differential_Calculus_(Feldman_Rechnitzer_and_Yeager)/03%3A_Derivatives/3.04%3A_Arithmetic_of_Derivatives_-_a_Differentiation_ToolboxSo far, we have evaluated derivatives only by applying Definition 2.2.1 to the function at hand and then computing the required limits directly. It is quite obvious that as the function being differen...So far, we have evaluated derivatives only by applying Definition 2.2.1 to the function at hand and then computing the required limits directly. It is quite obvious that as the function being differentiated becomes even a little complicated, this procedure quickly becomes extremely unwieldy.
- https://math.libretexts.org/Bookshelves/Calculus/CLP-1_Differential_Calculus_(Feldman_Rechnitzer_and_Yeager)/03%3A_Derivatives/3.11%3A_Implicit_DifferentiationImplicit differentiation is a simple trick that is used to compute derivatives of functions either when you don't know an explicit formula for the function, but you know an equation that the function...Implicit differentiation is a simple trick that is used to compute derivatives of functions either when you don't know an explicit formula for the function, but you know an equation that the function obeys or even when you have an explicit, but complicated, formula for the function, and the function obeys a simple equation.
- https://math.libretexts.org/Bookshelves/Calculus/CLP-1_Differential_Calculus_(Feldman_Rechnitzer_and_Yeager)/06%3A_Appendix/6.04%3A_D-_Hints_for_Exercises
- https://math.libretexts.org/Bookshelves/Calculus/CLP-1_Differential_Calculus_(Feldman_Rechnitzer_and_Yeager)/06%3A_Appendix/6.01%3A_A-_High_School_Material/6.1.10%3A_A.10_AreasArea of a rectangle \begin{align*} A &= b h \end{align*} Area of a triangle \begin{align*} A &= \frac{1}{2} b h = \frac{1}{2} ab \sin \theta \end{align*} Area of a circle \begin{align*} A &= \pi r^2 \...Area of a rectangle \begin{align*} A &= b h \end{align*} Area of a triangle \begin{align*} A &= \frac{1}{2} b h = \frac{1}{2} ab \sin \theta \end{align*} Area of a circle \begin{align*} A &= \pi r^2 \end{align*} Area of an ellipse \begin{align*} A &= \pi ab \end{align*}
- https://math.libretexts.org/Bookshelves/Calculus/CLP-1_Differential_Calculus_(Feldman_Rechnitzer_and_Yeager)/06%3A_Appendix/6.03%3A_C-_Root_FindingSuppose that we are given some function f(x) and we have to find solutions to the equation f(x)=0\text{.} To be concrete, suppose that f(x) = 8x^3+12x^2+6x-15\text{.} How do we go about so...Suppose that we are given some function f(x) and we have to find solutions to the equation f(x)=0\text{.} To be concrete, suppose that f(x) = 8x^3+12x^2+6x-15\text{.} How do we go about solving f(x)=0\text{?} To get a rough idea of the lay of the land, sketch the graph of f(x)\text{.} First observe that
