[Calculus] #3 Derivatives

문연수·2022년 8월 12일
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Calculus

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1. Definition of Derivative

The derivative of the function f(x)f(x) with respect to the variable xx is the function ff' whose value at xx is

f(x)=limh0f(x+h)f(x)hf'(x) = \lim\limits_{h \to 0}{\cfrac{f(x + h) - f(x)}{h}}

- Derivative Notation

f(x)=y=dydx=dfdx=ddxf(x)=D(f)(x)=Dxf(x)f'(x) = y' = \cfrac{dy}{dx}=\cfrac{df}{dx}=\cfrac{d}{dx}f(x)=D(f)(x)=D_{x}f(x)

The symbols d/dx and D indicate the operation of differentiation

2. Differentiability

If ff is continuous at x=ax = a, the left-hand derivative at x=ax = a is equal to the right-hand derivative at x=ax = a.

limh0f(a+h)f(a)h=limh0+f(a+h)f(a)h\lim\limits_{h \to 0^-}{\cfrac{f(a + h) - f(a)}{h} = \lim\limits_{h \to 0^+}{\cfrac{f(a + h) - f(a)}{h}}}
then, $f(x)$ is differentialble at $x = a$.

For the tangent slope to exist at x=ax=a, the function must be continuous, and the left-hand slope must equal to the right-hand slope.

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3. Power Rule

If nn is any real number, ddx(xn)=nxn1\cfrac{d}{dx}(x^n)=nx^{n-1}

 증명은 (x+h)n(x + h)^nnCr{}_nC{}_r 로 풀어쓴 뒤에 분자를 hh 로 나누면 h0h \to 0 일때 nxn1nx^{n-1} 만 남음.

4. Product Rule

If ff and gg are differentiable at xx, then fgf \circ g is differentiable and

ddx(f(x)g(x))=f(x)g(x)+f(x)g(x)\cfrac{d}{dx}(f(x) \circ g(x)) = f'(x)g(x) + f(x)g'(x)

증명은 f(x)g(x+h)f(x)g(x+h) 를 더하고 빼서 두 극한을 수렴하게 만들고 식을 정리하면 됨.

5. Quotient Rule

If ff and gg are differentialble at xx, then fg\cfrac{f}{g} is differentiable and

ddx(f(x)g(x))=f(x)g(x)+f(x)g(x)g2(x)\cfrac{d}{dx}{\left(\cfrac{f(x)}{g(x)}\right)} = \cfrac{f'(x)g(x) + f(x)g'(x)}{g^{2}(x)}

증명은 분모와 분자에 g(x+h)g(x)g(x+h)g(x) 를 곱하면 4. Product Rule 과 동일한 형태가 나옴.

6. Chain Rule

If f(u)f(u) is differentiable at the point u=g(x)u=g(x) and g(x)g(x) is differentiable at xx, then the composite function (fg)(x)=f(g(x))(f \circ g)(x) = f(g(x)) is differentiable at xx, and

(fg)(x)=f(g(x))g(x).(f \circ g)'(x) = f'(g(x)) \circ g'(x).

In Leibniz's notation, if y=f(u)y = f(u) and u=g(x)u = g(x), then

dyx=dydududx\cfrac{dy}{x} = \cfrac{dy}{du} \bullet \cfrac{du}{dx}

where dx/du is evaluated u=g(x)u = g(x)


Chain Rule 증명은 필자가 작성한 다음의 글을 참조.

7. Implicit Differentiation

When we cannot put an equation F(x,y)=0F(x, y) = 0 in the form y=f(x)y = f(x) to differentiable it in the usual way, we may still be able to find dy/dx by implicit differentiation:

 1. Differentiation both sizes of the equation with respect to xx treating yy as a differentiation function of xx
 2. Collect the terms with dy/dx on one side of the equation and solve the for dy/dx.

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8. Higher Order Derivatives

  • First Derivative    : y=dydxy'=\cfrac{dy}{dx}
  • Second derivative  : y=dydx=ddx(dydx)=d2ydx2y''=\cfrac{dy'}{dx}=\cfrac{d}{dx}\left(\cfrac{dy}{dx}\right)=\cfrac{d^2y}{dx^2}
    ...

이하 생략

9. Derivatives of Trigonometric Functions

ddx(sinx)=cosxddx(cosx)=sinxddx(tanx)=sec2xddx(cotx)=csc2xddx(secx)=secxtanxddx(cscx)=cscxcotx\cfrac{d}{dx}(\sin{x}) = \cos{x} \qquad\qquad \cfrac{d}{dx}(\cos{x})=-\sin{x} \newline \cfrac{d}{dx}(\tan{x}) = \sec^{2}{x} \qquad\qquad \cfrac{d}{dx}(\cot{x})=-\csc^{2}{x} \newline \cfrac{d}{dx}(\sec{x}) = \sec{x}\tan{x} \qquad\qquad \cfrac{d}{dx}(\csc{x})=-\csc{x}\cot{x}

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10. Derivatives of Exponential Functions

ddx(ex)=exddx(ef(x))=ef(x)f(x)ddx(ax)=axlna(a1,a>0)ddx(af(x))=af(x)f(x)lna(a1,a>0)\cfrac{d}{dx}(e^x) = e^x \qquad\qquad \cfrac{d}{dx}(e^{f(x)})=e^{f(x)} \bullet f'(x) \newline \cfrac{d}{dx}(a^x) = a^x \bullet \ln{a} \quad(a \ne 1, a>0)\qquad\qquad \cfrac{d}{dx}(a^{f(x)})=a^{f(x)} \bullet f'(x)\bullet \ln{a} \quad(a\ne1, a>0)\newline

11. Derivatives of Logarithmic Functions

ddx(lnx)=1xddx(logax)=1x1lnaddx(lnf(x))=1f(x)f(x)ddx(logaf(x))=1f(x)f(x)1lna\cfrac{d}{dx}(\ln{x})=\cfrac{1}{x} \qquad\qquad \cfrac{d}{dx}(\log_{a}{x}) = \cfrac{1}{x} \bullet \cfrac{1}{\ln{a}} \newline \cfrac{d}{dx}(\ln{f(x)})=\cfrac{1}{f(x)} \bullet f'(x) \qquad\qquad \cfrac{d}{dx}(\log_{a}{f(x)}) = \cfrac{1}{f(x)} \bullet f'(x) \bullet \cfrac{1}{\ln{a}}

12. Inverse Trigonometric Functions

y=sin1x=arcsinxy=cos1x=arccosxy=tan1x=arctanxy=csc1xy=sec1xy=cot1xy = \sin^{-1}{x} = \arcsin{x} \newline y = \cos^{-1}{x} = \arccos{x} \newline y = \tan^{-1}{x} = \arctan{x} \newline y = \csc^{-1}{x} \qquad y = \sec^{-1}{x} \qquad y = \cot^{-1}{x}

13. Derivatives of Inverse Trigonometric Functions

ddx(sin1x)=11x2ddx(cos1x)=11x2ddx(tan1x)=11+x2ddx(cot1x)=11+x2ddx(sec1x)=1xx21ddx(csc1x)=1xx21\cfrac{d}{dx}(\sin^{-1}{x}) = \cfrac{1}{\sqrt{1 - x^2}} \qquad\qquad \cfrac{d}{dx}(\cos^{-1}{x}) = \cfrac{-1}{\sqrt{1 - x^2}} \newline \cfrac{d}{dx}(\tan^{-1}{x}) = \cfrac{1}{1 + x^2} \qquad\qquad \cfrac{d}{dx}(\cot^{-1}{x}) = \cfrac{-1}{1 + x^2} \newline \cfrac{d}{dx}(\sec^{-1}{x}) = \cfrac{1}{|x|\sqrt{x^2 - 1}} \qquad\qquad \cfrac{d}{dx}(\csc^{-1}{x}) = \cfrac{-1}{|x|\sqrt{x^2 -1}} \newline

 주의해야 할 것은 역함수 내의 식이 복잡할 때 yyxx 를 그냥 치환하게 되면 문제가 생긴다. 반드시 서로 다른 함수로 인지하고 Chain Rule 을 통해 풀어서 미분해야 한다.

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14. Derivatives of Inverse Functions

Let ff be differentiable and a one-to-one function on a domain.

ddx[f1(x)]=(f1)(x)=1f(f1(x))\cfrac{d}{dx}[f^{-1}(x)] = (f^{-1})'(x) = \cfrac{1}{f'(f^{-1}(x))}

증명은 f(f1(x))=xf(f^{-1}(x)) = x 이라는 사실을 통해 Chain Rule 을 통해 풀어서 식을 정리하면 된다.

출처

[책] Man Sik Min · Hyeong Chul Jeong · Hyejung Lee, 『CALCULUS』, 한티미디어

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