[Note] The Derivative science

Derivative

Conductivity can be determined

  1. $ f (x) $ in $ $ x_0 derivable == == conditions sufficient: $ f (x) $ $ x_0 about $ F may be turned and has $ ' - (X) = F' + (X) $
  2. Must be turned continuously, it is not necessarily continuous derivable

Chain rule derivation

设$y=f(x)$,则$f'(t)=\frac{{\rm d}y}{{\rm d}x}=\frac{{\rm d}y}{{\rm d}t}\frac{{\rm d}t}{{\rm d}x}$

Used in the derivation of the complex composite function


$eg.1求(a^x)'$
$a^x=e^{xlna}\rightarrow 令u=xlna \Rightarrow$
$$
\frac{{\rm d}a^x}{{\rm d}x}=\frac{{\rm d}e^u}{{\rm d}u}\frac{{\rm d}u}{{\rm d}x}=e^ulna=a^xlna
$$


$ eg.2 provided y = sin \ sqrt {x ^ 2 + x + 1}, seeking Y '$
$ set u = x ^ 2 + x + 1, v = \ sqrt {u}, by the chain rule \ Rightarrow $
$$ \ FRAC {{\ RM Y} {D} {\ RM D} X} = \ FRAC {{\ RM Y} {D} {\ RM V D}} \ FRAC {{\ RM V D}} {{\ rm d} u} \ frac {{\ rm d} u} {{\ rm d} x} = cosv \ frac {1} {2 \ sqrt {u}} (2x + 1) = \ frac { 2x + 1} {2 \ sqrt {x ^ 2 + x + 1}} cos \ sqrt {x ^ 2 + x + 1} $$


Logarithmic derivative

$y=f(x)\rightarrow lny=lnf(x)\rightarrow (lny)'=(lnf(x))'\Rightarrow$
$y'\frac{1}{y}=(lnf(x))'\Rightarrow y'=y(lnf(x))'$

$eg.3设y=\sqrt[3]{\frac{x^2(x-1)(x-2)}{(x+1)(x+3)^2}},求y'$
$\begin{aligned}
y' & =y(ln\frac{x^2(x-1)(x-2)}{(x+1)(x+3)^2})' \\
& = y(\frac{2}{3}lnx+\frac{1}{3}ln(x-1)+\frac{1}{3}ln(x+2)-\frac{1}{3}ln(x+1)-\frac{2}{3}ln(x+3))' \\
& = \sqrt[3]{\frac{x^2(x-1)(x-2)}{(x+1)(x+3)^2}}(\frac{2}{3(x^2+1)}+\frac{2}{x(x+3)}+\frac{1}{3(x+2)}) \\
\end{aligned}
$

Function of the logarithm of the complex compound used in derivation function contains a power function, multiplication, division

Inverse function derivation

$由链式法则可得:\frac{{\rm d}y}{{\rm d}x}=\frac{1}{\frac{{\rm d}x}{{\rm d}y}}$
$eg.4设x=arcsiny,求x'$
$x=arcsiny\rightarrow y=sinx\Rightarrow $
$$
\frac{{\rm d}x}{{\rm d}y} = \frac{1}{\frac{{\rm d}y}{{\rm d}x}} = \frac{1}{\frac{{\rm d}sinx}{{\rm d}x}} = \frac{1}{cosx} = \frac{1}{\sqrt{1-sin^2x}}=\frac{1}{\sqrt{1-y^2}}
$$


Derivation of power exponential function

Into a first end of a constant, then the derivative of the index

$ EG.5 setting request y = x ^ x Y '$
$ Y' = (E ^ {} XLNX) '= (XLNX)' XLNX} = E {^ (LNX +. 1) X ^ X $


Order derivative

$ (sinx) ^ {(n)} = sin (\ frac {n \ pi} {2} + x) $
$ (cosx) ^ {(n)} = cos (\ frac {n \ pi} {2} + x) $
$ ln (1 + x) ^ {(n)} = \ frac {(- 1) ^ {n-1} (n-1)!} {(1 + x) ^ n} $
$ sine (ax + b) ^ {(n)} = a ^ nsin (ax + b + \ frac {n \ pi} {2}) $
$ (e ^ {ax + b}) ^ {(n)} = a ^ ne ^ {ax + b} $
$ [u (x) v (x)] ^ {(n)} = \ sum_ {k = 0} ^ {n} C_ {n} ^ {k} u ^ {(k )} ^ {v (nk)} $


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Origin www.cnblogs.com/bbqub/p/11970730.html