# Waves/Derivatives

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### Math Tutorial -- Derivatives 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 $x$ :

$g(x)={\frac {dy(x)}{dx}}.$ (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 Derivatives

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

Table of Derivatives
${d \over dx}c=0$
${d \over dx}x=1$
${d \over dx}cx=c$
${d \over dx}|x|={x \over |x|}=\operatorname {sgn} x,\qquad x\neq 0$
${d \over dx}x^{c}=cx^{c-1}$  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
${d \over dx}e^{x}=e^{x}$
${d \over dx}\log _{c}x={1 \over x\ln c}$  c > 0, c ≠ 1
${d \over dx}\ln x={1 \over x}$
${d \over dx}\sin x=\cos x$
${d \over dx}\cos x=-\sin x$
${d \over dx}\tan x=\sec ^{2}x$
${d \over dx}\sec x=\tan x\sec x$
${d \over dx}\cot x=-\csc ^{2}x$
${d \over dx}\csc x=-\csc x\cot x$
${d \over dx}\arcsin x={1 \over {\sqrt {1-x^{2}}}}$
${d \over dx}\arccos x={-1 \over {\sqrt {1-x^{2}}}}$
${d \over dx}\arctan x={1 \over 1+x^{2}}$
${d \over dx}\operatorname {arcsec} x={1 \over |x|{\sqrt {x^{2}-1}}}$
${d \over dx}\operatorname {arccot} x={-1 \over 1+x^{2}}$
${d \over dx}\operatorname {arccsc} x={-1 \over |x|{\sqrt {x^{2}-1}}}$
${d \over dx}\sinh x=\cosh x$
${d \over dx}\cosh x=\sinh x$
${d \over dx}\tanh x=\operatorname {sech} ^{2}x$
${d \over dx}\operatorname {sech} x=-\tanh x\operatorname {sech} x$
${d \over dx}\operatorname {coth} x=-\operatorname {csch} ^{2}x$
${d \over dx}\operatorname {csch} x=-\operatorname {coth} x\operatorname {csch} x$
${d \over dx}\operatorname {arsinh} x={1 \over {\sqrt {x^{2}+1}}}$
${d \over dx}\operatorname {arcosh} x={1 \over {\sqrt {x^{2}-1}}}$
${d \over dx}\operatorname {artanh} x={1 \over 1-x^{2}}$
${d \over dx}\operatorname {arsech} x={1 \over x{\sqrt {1-x^{2}}}}$
${d \over dx}\operatorname {arcoth} x={1 \over 1-x^{2}}$
${d \over dx}\operatorname {arcsch} x={-1 \over |x|{\sqrt {1+x^{2}}}}$

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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