The derivative of the function f(x) with respect to the variable x is the function f′ whose value at x is
f′(x)=h→0limhf(x+h)−f(x)
- Derivative Notation
f′(x)=y′=dxdy=dxdf=dxdf(x)=D(f)(x)=Dxf(x)
The symbols d/dx and D indicate the operation of differentiation
2. Differentiability
If f is continuous at x=a, the left-hand derivative at x=a is equal to the right-hand derivative at x=a.
h→0−limhf(a+h)−f(a)=h→0+limhf(a+h)−f(a)
then, $f(x)$ is differentialble at $x = a$.
For the tangent slope to exist at x=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 n is any real number, dxd(xn)=nxn−1
증명은 (x+h)n 을 nCr 로 풀어쓴 뒤에 분자를 h 로 나누면 h→0 일때 nxn−1 만 남음.
4. Product Rule
If f and g are differentiable at x, then f∘g is differentiable and
dxd(f(x)∘g(x))=f′(x)g(x)+f(x)g′(x)
증명은 f(x)g(x+h) 를 더하고 빼서 두 극한을 수렴하게 만들고 식을 정리하면 됨.
5. Quotient Rule
If f and g are differentialble at x, then gf is differentiable and
dxd(g(x)f(x))=g2(x)f′(x)g(x)+f(x)g′(x)
증명은 분모와 분자에 g(x+h)g(x) 를 곱하면 4. Product Rule 과 동일한 형태가 나옴.
6. Chain Rule
If f(u) is differentiable at the point u=g(x) and g(x) is differentiable at x, then the composite function (f∘g)(x)=f(g(x)) is differentiable at x, and
When we cannot put an equation F(x,y)=0 in the form 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 x treating y as a differentiation function of x
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′=dxdy
Second derivative : y′′=dxdy′=dxd(dxdy)=dx2d2y
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