# Waves/Derivatives

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### Math Tutorial -- DerivativesEdit

Figure 1.15: Estimation of the derivative, which is the slope of the tangent line. When point B approaches point A, the slope of the line AB approaches the slope of the tangent to the curve at point A.

This section provides a quick introduction to the idea of the derivative. For a more detailed discussion and exploration of the differentiation and of Calculus, see Calculus and Differentiation.

Often we are interested in the slope of a line tangent to a function $y(x)$ at some value of $x$. This slope is called the derivative and is denoted $dy/dx$. Since a tangent line to the function can be defined at any point $x$, the derivative itself is a function of $x$:

$g(x) = \frac{d y(x)}{dx} .$ (2.25)

As figure 1.15 illustrates, the slope of the tangent line at some point on the function may be approximated by the slope of a line connecting two points, A and B, set a finite distance apart on the curve:

$\frac{dy}{dx} \approx \frac{\Delta y}{\Delta x} .$ (2.26)

As B is moved closer to A, the approximation becomes better. In the limit when B moves infinitely close to A, it is exact.

## Table of DerivativesEdit

Derivatives of some common functions are now given. In each case $a$ is a constant.

Table of Derivatives
${d \over dx} c = 0$
${d \over dx} x = 1$
${d \over dx} cx = c$
${d \over dx} |x| = {x \over |x|} = \sgn x,\qquad x \ne 0$
${d \over dx} x^c = cx^{c-1}$ where both xc and cxc-1 are defined.
${d \over dx} \left({1 \over x}\right) = {d \over dx} \left(x^{-1}\right) = -x^{-2} = -{1 \over x^2}$
${d \over dx} \left({1 \over x^c}\right) = {d \over dx} \left(x^{-c}\right) = -{c \over x^{c+1}}$
${d \over dx} \sqrt{x} = {d \over dx} x^{1\over 2} = {1 \over 2} x^{-{1\over 2}} = {1 \over 2 \sqrt{x}}$ x > 0
${d \over dx} c^x = {c^x \ln c}$ c > 0[/itex]
${d \over dx} e^x = e^x$
${d \over dx} \log_c x = {1 \over x \ln c}$ c > 0, $c \ne 1$
${d \over dx} \ln x = {1 \over x}$
${d \over dx} \sin x = \cos x$
${d \over dx} \cos x = -\sin x$
${d \over dx} \tan x = \sec^2 x$
${d \over dx} \sec x = \tan x \sec x$
${d \over dx} \cot x = -\csc^2 x$
${d \over dx} \csc x = -\csc x \cot x$
${d \over dx} \arcsin x = { 1 \over \sqrt{1 - x^2}}$
${d \over dx} \arccos x = {-1 \over \sqrt{1 - x^2}}$
${d \over dx} \arctan x = { 1 \over 1 + x^2}$
${d \over dx} \arcsec x = { 1 \over |x|\sqrt{x^2 - 1}}$
${d \over dx} \arccot x = {-1 \over 1 + x^2}$
${d \over dx} \arccsc x = {-1 \over |x|\sqrt{x^2 - 1}}$
${d \over dx} \sinh x = \cosh x$
${d \over dx} \cosh x = \sinh x$
${d \over dx} \tanh x = \mbox{sech}^2 x$
${d \over dx} \mbox{sech} x = - \tanh x \mbox{sech} x$
${d \over dx} \mbox{coth} x = - \mbox{csch}^2 x$
${d \over dx} \mbox{csch} x = - \mbox{coth} x \mbox{csch} x$
${d \over dx} \mbox{arcsinh} x = { 1 \over \sqrt{x^2 + 1}}$
${d \over dx} \mbox{arccosh} x = { 1 \over \sqrt{x^2 - 1}}$
${d \over dx} \mbox{arctanh} x = { 1 \over 1 - x^2}$
${d \over dx} \mbox{arcsech} x = { 1 \over x\sqrt{1 - x^2}}$
${d \over dx} \mbox{arccoth} x = { 1 \over 1 - x^2}$
${d \over dx} \mbox{arccsch} x = {-1 \over |x|\sqrt{1 + x^2}}$

The product and chain rules are used to compute the derivatives of complex functions. For instance,

$\frac{d}{dx} ( \sin (x) \cos (x)) = \frac{d \sin (x)}{dx} \cos (x) + \sin (x) \frac{d \cos (x)}{dx} = \cos^2 (x) - \sin^2 (x)$

and

$\frac{d}{dx} \log ( \sin (x) ) = \frac{1}{\sin (x)} \frac{d \sin (x)}{dx} = \frac{\cos (x)}{\sin (x)} .$

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