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If you wish to make an apple pie from scratch, you must first invent the universe. — Carl Sagan

Definition. Suppose that a curve is given as the graph of a function, y = f(x). The tangent line is the straight line that just touches the curve of the given function at one point P = (x0, y0) -1.a.-

The tangent line’s equation at a point (x0, y0) is y - y0 = m (x - x0), y0 = f(x0). The slope m of a tangent line at a point P(x0, y0) is its derivative at that point, m = f’(x0). In other words, the derivative of a function at a given point f’(x0) is the slope of the tangent line at that point.   The slope of the tangent line is the limit of the slopes of the secant lines PQ as they approach the tangent line or as Q approach P (P is obviously fixed), that is -1.b., 1.c.-,

m = f’(x0) = $\lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x}$ = $\lim_{\Delta x \to 0} \frac{f(x_0+\Delta x) -f(x_0)}{\Delta x}$

Example: f(x)=1x. 1.d.

$\frac{f(x_0+\Delta x) -f(x_0)}{\Delta x} = \frac{\frac{1}{x_0+\Delta x} -\frac{1}{x_0}}{\Delta x} = \frac{1}{\Delta x} \frac{x_0-x_0-\Delta x}{(x_0+\Delta x)x_0} = \frac{-1}{(x_0+\Delta x)x_0}$

And therefore, $\lim_{\Delta x \to 0} \frac{f(x_0+\Delta x) -f(x_0)}{\Delta x} = \lim_{\Delta x \to 0}\frac{-1}{(x_0+\Delta x)x_0} = \frac{-1}{x_0^{2}}$

Observe that the slope is negative and as x0 → ∞, the line tangent has less and less sleep

Let’s find the areas of triangles enclosed by the axes and the tangent line to y = 1x.

The tangent line equation is y - y0 = -1x02 (x - x0).

The x-intercept is the point where the tangent line crosses the x-axis, that is, y = 0.

y - y0 = 0 - y0 = 0 - 1x0 = -1x02 (x - x0) = -xx02+1x0

$\frac{x}{x_0^{2}}=\frac{2}{x_0}$ ⇨ x = 2x0. The x-intercept is (2x0, 0).

The y-intercept is the point where the tangent line crosses the y-axis, that is, x = 0. Because of symmetry (y = 1x ↔ xy = 1 ↔ x = 1y), the y-intercept is (0, 2y0).

Finally, the area of a triangle is defined as the region enclosed by it three sides. It is equal to half of the base times height, i.e., A = 12bh=12(2x0)(2y0) = 2. Note that y0 = 1x0.

Let’s calculate f’(x)=xn, n = 1, 2, 3, ….

We normally do not use x0, just x, and consider that x is fixed, but $\Delta x$ moves towards zero.

$\frac{f(x+\Delta x) -f(x)}{\Delta x} = \frac{(x+\Delta x)^{n} -x^{n}}{\Delta x} = \frac{x^{n}+n\Delta xx^{n-1} + O((\Delta x)^{2}) -x^{n}}{\Delta x} = nx^{n-1} + O(\Delta x)$

$f’(x^{n})=\lim_{\Delta x \to 0}nx^{n-1} + O(\Delta x) = nx^{n-1}$

If y is a function of x, y = f(x), a change in x from x1 to x2 is denoted by $\Delta x = x_2 - x_1$, and the corresponding change in y is denoted by $\Delta y = f(x_2) - f(x_1)$. The difference quotient $\frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$ is called the average rate of change of y with respect to x. This is the slope of the line segment PQ, where P(x1, f(x1) and Q(x2, f(x2)). The instantaneous rate of change of y with respect to x is the limit of the slopes of line segments PQ as Q gets closer and closer to P, that is,

$lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x} = lim_{x_2 \to x_1}\frac{f(x_2) - f(x_1)}{x_2 - x_1} = f’(x_1).$

Therefore, the derivate f'(a) is the instantaneous rate of change of y = f(x) with respect to x when x = a. Some examples are:

Example: If the distance that a person covers given at a certain time is d(t)=t2 -t +2. Find the average speed for the first five seconds and the instantaneous speed after two seconds.

$v_m = \frac{d(t=5) - d(t=0)}{5} = \frac{22 - 2 m}{5s} = 4 m/s.$

$v = lim_{\Delta t \to 0}\frac{\Delta s}{\Delta t} = \frac{ds}{dt} = 2t - 1, v(2) = d’(2)= 3 m/s.$

Derivatives of Trigonometric Functions

$\frac{sin(x+\Delta x) -sin(x)}{\Delta x} = \frac{sin(x)cos(\Delta x) + cos(x)sin(\Delta x) -sin(x)}{\Delta x}$ [ by sin(a+b) = sin(a)cos(b) + cos(a)sin(b)] = $sin(x)(\frac{cos(\Delta x)-1}{\Delta x}) + cos(x)(\frac{sin(\Delta x)}{\Delta x})$

$lim_{\Delta x \to 0} \frac{sin(x+\Delta x) -sin(x)}{\Delta x} = lim_{\Delta x \to 0} sin(x)(\frac{cos(\Delta x)-1}{\Delta x}) + cos(x)(\frac{sin(\Delta x)}{\Delta x}) = $

And by Limits, $\lim_{x \to 0} (\frac{sin(x)}{x}) = 1$ and $\lim_{x \to 0} (\frac{1-cos(x)}{x}) = 0$ and limit laws,

$ = (-1)·0·sin(x) + cos(x)*1 = cos(x) $

$\frac{cos(x+\Delta x) -cos(x)}{\Delta x} = \frac{cos(x)cos(\Delta x) - sin(x)sin(\Delta x) -cos(x)}{\Delta x}$ [ by cos(a+b) = cos(a)cos(b) - sin(a)sin(b)] = $cos(x)(\frac{cos(\Delta x)-1}{\Delta x}) - sin(x)(\frac{sin(\Delta x)}{\Delta x})$

And by Limits, $\lim_{x \to 0} (\frac{sin(x)}{x}) = 1$ and $\lim_{x \to 0} (\frac{1-cos(x)}{x}) = 0$ and limit laws,

$ = cos(x)·-1·0 -sin(x)·1 = -sin(x).$

$\frac{d}{dx}\frac{sinx}{cosx}$ =[Quotient Rule] $\frac{cosxcosx+sinxsinx}{cos^{2}x} = \frac{1}{cos^{2}x}$

General rules

n = 0, u(x) = 0. $\lim_{\Delta x \to 0}\frac{0 -0}{\Delta x} = 0.$

Proof by induction. n = 1, u(x) = x. $\lim_{\Delta x \to 0}\frac{x+\Delta x-x}{\Delta x} = 1.$ ⇨ u’(x) = 1.

Suppose it holds for n-1. u(x)=xn = x·xn-1 ⇨ [Product rule] u’(x)=(xn-1)·1+(n-1)xn-2·x= xn-1 + (n-1)xn-1 = n·xn-1

If u(x)=xn, n≥0, then u’(x)=nxn-1.

Let u(x)=1, v(x)=xn ⇨ [Quotient rule] $ (\frac{1}{x^{n}})’ = \frac{u’v - uv’}{v^{2}} = \frac{-nx^{n-1}}{x^{2n}} = -nx^{-n-1}$

If u(x)=xn, n=0, ±1, ±2, ±3,… then u’(x)=nxn-1.

$\frac{u(x+\Delta x)v(x+\Delta x) -u(x)v(x)}{\Delta x} = \frac{(u+\Delta u)(v+\Delta v) -uv}{\Delta x} = $

Notice that △u and △v are the increments of u and v respectively, and x has been omitted for brevity.

$ = \frac{uv+u\Delta v+v\Delta u+\Delta u\Delta v -uv}{\Delta x} = u\frac{\Delta v}{\Delta x} + v\frac{\Delta u}{\Delta x} + \frac{\Delta u\Delta v}{\Delta x}$

(uv)’ = $\lim_{\Delta x \to 0} u\frac{\Delta v}{\Delta x} + v\frac{\Delta u}{\Delta x} + \frac{\Delta u\Delta v}{\Delta x} = \lim_{\Delta x \to 0} u\frac{\Delta v}{\Delta x} + \lim_{\Delta x \to 0} v\frac{\Delta u}{\Delta x} + \lim_{\Delta x \to 0} \frac{\Delta u\Delta v}{\Delta x} = u\lim_{\Delta x \to 0} \frac{\Delta v}{\Delta x} + v\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} + \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x}\lim_{\Delta x \to 0} \Delta v = $ […]

$\lim_{\Delta x \to 0} \Delta v = \lim_{\Delta x \to 0} \frac{\Delta v}{\Delta x}·\Delta x = lim_{\Delta x \to 0} \frac{\Delta v}{\Delta x}·lim_{\Delta x \to 0}\Delta x = v’·0 = 0$

[…] = u·v’ + v.u’ + 0 = u·v’ + v.u'.

$(\frac{u}{v})’ = \frac{u’v - uv’}{v^{2}}$

$\frac{\frac{u(x+\Delta x)}{v(x+\Delta x)} -\frac{u(x)}{v(x)}}{\Delta x} = \frac{\frac{u+\Delta u}{v+\Delta v} -\frac{u}{v}}{\Delta x} = \frac{\frac{u·v+\Delta u·v-u·v-u\Delta v}{(v+\Delta v)v}}{\Delta x} = \frac{\frac{\Delta u·v-u\Delta v}{(v+\Delta v)v}}{\Delta x} = \frac{\frac{\Delta u·v-u\Delta v}{\Delta x}}{(v+\Delta v)v} = \frac{\frac{\Delta u}{\Delta x}v -u\frac{\Delta v}{\Delta x}}{(v+\Delta v)v} $

$\lim_{\Delta x \to 0} \frac{\frac{\Delta u}{\Delta x}v -u\frac{\Delta v}{\Delta x}}{(v+\Delta v)v} = \frac{u’v-uv’}{v^{2}}$ [We have already proven that $\lim_{\Delta x \to 0} \Delta v = 0$]

Let t = v(x), $\frac{\Delta u}{\Delta x} = \frac{\Delta u}{\Delta t}\frac{\Delta t}{\Delta x}$. And therefore, △x→0, $\frac{du}{dx} =\frac{du}{dt}\frac{dt}{dx}$

Example: y = cos2x, y’= 2·cos(x)·(-sin(x)) = -2cos(x)sin(x).

y = sin(nx), y’=cos(nx)·n = n·cos(nx).

Higher Order Derivatives.

Because the derivative of a function u = u(x) is itself a function, you can take its derivate, which is generally referred to as the second derivative of u and written as u"(x), u2(x), or even simply u". This differentiation process can be continued to find the third, fourth, and so on, which are called higher order derivatives of u.

Example: u = sin(x), u’ = cos(x), u" = -sin(x), u3 = -cos(x), u4 = sin(x).

Other notations: u’= $\frac{du}{dx}=\frac{d}{dx}u$. This is the differential operator. It is applied to a function.

u" = $\frac{d}{dx}\frac{du}{dx}=\frac{d}{dx}\frac{d}{dx}u = (\frac{d}{dx})^{2}u = \frac{d^{2}u}{dx^{2}}.$

u3 = $\frac{d^{3}u}{dx^{3}} = D^{3}u.$

$D^{m}x^{n}.$ Let m = 1, $D x^{n}=nx^{n-1}.~ D^{2} x^{n}=n(n-1)x^{n-2}.~ D^{3} x^{n}=n(n-1)(n-2)x^{n-3}.~ D^{n} x^{n}=n!.~ D^{n+1} x^{n}=0.$

Implicit Differentiation

y = xm/n ↔ yn = xm. Let’s apply the differential operator ddx.

$\frac{d}{dx}y^{n} = \frac{d}{dx}x^{m} = mx^{m-1}$

$\frac{d}{dx}y^{n} = (\frac{d}{dy}y^{n})\frac{dy}{dx} = ny^{n-1}\frac{dy}{dx} = mx^{m-1} ⇨ \frac{dy}{dx} = \frac{m}{n} \frac{x^{m-1}}{y^{n-1}} = \frac{m}{n} \frac{x^{m-1}}{(x^\frac{m}{n})^{n-1}} = ax^{m-1-(n-1)\frac{m}{n}}$

$m-1-(n-1)\frac{m}{n} = m -1 -m + \frac{m}{n} = \frac{m}{n} -1 $

$\frac{dy}{dx} = ax^{a-1},~ where~ a=\frac{m}{n}$

y2 = 1 - x2 ⇨ $y=\pm\sqrt{1-x^{2}}$ (Explicit definition). Let’s take the positive branch, $y=+\sqrt{1-x^{2}}=(1-x^{2})^{\frac{1}{2}}$

y’ = $\frac{1}{2}(1-x^{2})^{\frac{-1}{2}}(-2x) = -x(1-x^{2})^{\frac{-1}{2}}=\frac{-x}{\sqrt{1-x^{2}}}$

Or alternatively, x2 + y2 = 1 ⇨ $\frac{d}{dx}(x^{2}+y^{2}=1)~⇨~2x+2yy’=0~⇨~y’=\frac{-x}{y}.$ The implicit way does not need to take only one branch.

Explicit. $y^{2}=\frac{-x\pm\sqrt{x^{2}+8}}{2}, y=\pm\sqrt{\frac{-x\pm\sqrt{x^{2}+8}}{2}}$. It is not a good approach.

Implicit. 4y3y’+ y2 + 2yy’x = 0. ⇨ (4y3+2xy)y’ + y2 = 0 ⇨ y’ = $\frac{-y^{2}}{4y^{3}+2xy}$

Derivate of inverse functions

The inverse function of a function f is a function that undoes the operation of f.

$\frac{d}{dx}(tany=x)⇨~ \frac{d}{dy}tany\frac{dy}{dx}=1⇨~ \frac{1}{cos^{2}y}\frac{dy}{dx}=1⇨~ y’=cos^{2}y$

$(tan^{-1}x)’=\frac{d}{dx}\tan^{-1}x=cos^{2}(y)$ [by cosy=$\frac{1}{\sqrt{1+x^{2}}}$ 1.c.] = $\frac{1}{1+x^{2}}$

$\frac{d}{dx}siny=1⇨~ \frac{d}{dy}siny\frac{dy}{dx}=1⇨~ cosy\frac{dy}{dx}=1$

$y’= \frac{1}{cos(y)}=~ \frac{1}{\sqrt{1-sin^{2}y}}=~ \frac{1}{\sqrt{1-x^{2}}}$

$(sin^{-1}x)’=\frac{d}{dx}sin^{-1}x=~ \frac{1}{\sqrt{1-x^{2}}}$

$\frac{d}{dx}secx =~(-1)(cosx)^{-2}(-sin(x))=\frac{sin(x)}{cos(x)^{2}}=secx·tanx.$

Derivate of Exponential and Logarithms functions.

An exponential function is a function of the form f(x) = ax, where “x” is a variable and “a” is a constant which is called the base of the function and it should be a positive real number (a>0).  

am/n= $\sqrt[n]{a^{m}}$

f(x)=ax, let’s calculate $\frac{d}{dx}a^{x}$

f’(x) = $\frac{d}{dx}a^{x} =~ \lim_{\Delta x \to 0} \frac{a^{x+\Delta x}-a^{x}}{\Delta x} = \lim_{\Delta x \to 0}a^{x}\frac{a^{\Delta x}-1}{\Delta x}$ [by axay=ax+y] = $a^{x}\lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}$

Let’s define M(a) = $\lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}$. It is the slope of ax at x = 0 because $\frac{d}{dx}a^{x} =~ a^{0}M(a)=~ M(a)$

f’(x) = $\frac{d}{dx}a^{x} = M(a)a^{x}$

Definition. The number e, also known as Euler’s number, is a mathematical constant approximately equal to 2.71828 that can be characterized in many ways. It is the unique positive number 'a' such that the graph of the function y=ax has a slope of 1 at x = 0, that is, M(e)=1. -1.a.-

a = e, $\frac{d}{dx}a^{x} = M(a)a^{x} ⇨~ \frac{d}{dx}e^{x} = e^{x}$

Besides, M(e)=1, $\frac{d}{dx}e^{x} = e^{0}=1$ Besides, M(e)=1, $\frac{d}{dx}e^{x}|_{x=0} = e^{0}=1$

Let a=2, f(x)=2x. Then, f(kx)=2kx=(2k)x=bx where b=2k.

$\frac{d}{dx}b^{x} = \frac{d}{dx}f(kx)=kf’(kx)$

$\frac{d}{dx}b^{x}|_{x=0}=kf’(0)=kM(2)$ Therefore, b=e when k=$\frac{1}{M(2)}$

We call the inverse of ax the logarithmic function with base a, that is, logax=y ↔ ay=x. The logarithm with base e is the natural logarithm. The natural logarithm is the inverse of ex, that is, lnx=y ↔ ay=x, -1.b.-

Let w = lnx, $\frac{d}{dx}lnx$

ew = x, $\frac{d}{dx}e^{w}=\frac{d}{dw}e^{w}\frac{dw}{dx} = 1$

$\frac{d}{dx}e^{w}=e^{w}\frac{dw}{dx} = 1$ Therefore, $\frac{d}{dx}lnx=\frac{1}{x}$

$\frac{d}{dx}a^{x}=\frac{d}{dx}((e^{lna})^{x})=~ \frac{d}{dx}((e^{lna})^{x})=~ \frac{d}{dx}e^{xlna}=~ e^{xlna}lna=e^{xlna}·lna=a^{x}·lna$

$\frac{d}{dx}a^{x}=(lna)a^{x}$, so M(a)=lna

$(lnu)’=\frac{d}{dx}lnu = \frac{d}{du}lnu\frac{du}{dx} = \frac{1}{u}u’=\frac{u’}{u}$

Let’s calculate $\frac{d}{dx}a^{x}$ taking into consideration that u = ax, so lnu=x·lna, and therefore, (lnu)’=lna,

$(lnu)’=\frac{u’}{u}=lna⇨~ u’=u·lna⇨~ \frac{d}{dx}a^{x} = (lna)·a^{x}$

Exercise. u=xx, lnu=x·lnx ⇨ $(lnu)’=lnx + x\frac{1}{x} = lnx + 1 ⇨~ \frac{u’}{u}= lnx + 1⇨~ u’=x^{x}(lnx+1).$

Exercise. $\lim_{n \to \infin}(1+\frac{1}{n})^{n}$

$ln((1+\frac{1}{n})^{n}) = nln(1+\frac{1}{n})=\frac{1}{\Delta x}ln(1+\Delta x) = $ [where $\Delta x=\frac{1}{n}$] =$\frac{1}{\Delta x}(ln(1+\Delta x)-ln1)$ [ln1 = 0]

And therefore, $\Delta x→0$, $\frac{1}{\Delta x}(ln(1+\Delta x)-ln1)$ is $\frac{d}{dx} lnx|_ {x=1} =~ \frac{1}{x}|_{x=1} = 1$

$\lim_{n \to \infin}(1+\frac{1}{n})^{n} = e^{lim_{n \to \infin} ln((1+\frac{1}{n})^{n})}=e.$

Definition. The number e, also known as Euler’s number, is a mathematical constant approximately equal to 2.71828 that can be characterized in many ways. It is the unique positive number 'a' such that the graph of the function y=ax has a slope of 1 at x = 0, that is, M(e)=1, Fig. 1.a. It is also the limit of (1 + 1n)n as n approaches infinity. e = $\lim_{n \to \infin}(1+\frac{1}{n})^{n}.$

Power rule for real exponents. If u(x)=xr, then u’(x)=rxr-1, ∀r ∈ ℝ.

  1. $x^{r}=(e^{lnx})^{r}=e^{rlnx}$ ⇨ $\frac{d}{dx}x^{r}=\frac{r}{x}e^{rlnx}=\frac{r}{x}x^{r}=rx^{r-1}$
  2. $u=x^{r}, ~lnu=rlnx$ ⇨ $\frac{u’}{u}=\frac{r}{x}⇨~ u’=\frac{u}{x}=rx^{r-1}$


$\frac{d}{dx}e^{xtan^{-1}x} = e^{xtan^{-1}x}(tan^{-1}x + \frac{x}{1+x^{2}})$

$\lim_{u \to 0}\frac{e^{u}-1}{u} =~\frac{d}{du}e^{u}|_{u=0}=e^{0}=1.$

By taking into account that, e’(x)=$\lim_{\Delta x \to 0}\frac{e^{x+\Delta x}-e^{x}}{\Delta x}$, and therefore, e’(0) = $\lim_{\Delta x \to 0}\frac{e^{\Delta x}-1}{\Delta x}$

Linear and quadratic approximation

Linear approximation does a very good job of approximating values of f(x) as long as we stay near x = a, that is, in the neighborhood around a. If a function y = f(x) is differentiable at a point (a, f(a)), the function looks like its tangent line in a small open interval containing x=a. It allows us to approximate the original function f (it is sometimes computationally or algebraically complicated) with a simpler function L that is linear.

The tangent line is f(x) = f(x0) + f’(x0)(x-x0).

Let x0=1. Some examples are: lnx ≅ x - 1 (-1.a.-)

Let x0=0, f(x) = f(0) + f’(0)(x). Some examples are: sinx ≅ x (1.b), cosx ≅ 1 (1.c), ex ≅ 1+x, ln(1+x) ≅ x, (1+x)r ≅ 1 +rx.

Example. Find linear approximation near x = 0 of $\frac{e^{-3x}}{\sqrt{1+x}}$

$\frac{e^{-3x}}{\sqrt{1+x}} ≅ e^{-3x}(\sqrt{1+x})^{-1/2} ≅ (1-3x)(1-\frac{1}{2}x) ≅~ 1 -3x -\frac{1}{2}x +\frac{3}{2}x^{2} ≅ 1 -3x -\frac{1}{2}x ≅ 1 -\frac{7}{2}x$. Observe that we do not care about quadratic and higher other terms and consider them as negligible as long as we are close enough to 0.

When using approximation you sacrifice some accuracy for the ability to perform complex calculations easily. Quadratic approximation is used when the linear approximation is not enough. The basic formula for quadratic approximation with x0 is:

f(x) ≅ f(x0) + f’(x0)(x-x0) + f’’(x0)2(x-x0)2 (x ≅ x0)

It gives a best-fit parabola to a function. Ideally, the quadratic approximation of a quadratic function, say f(x) = a +bx +cx2 should be identical to the original function.

f(x) = a +bx +cx2
f’(x) = b + 2cx
f’’(x) = 2c

Set the base x0 as zero. Next, we try to recover the values of a, b, and c. f(0) = a
f’(0) = b
f’’(0) = 2c ⇒ c = f’’(0)2

f(x) ≅ f(x0) + f’(x0)(x-x0) + f’’(x0)2(x-x0)2 (x ≅ x0, say x = 0) = f(0) + f’(0)(x) + f’’(0)2x2 (x ≅ 0) ≅ a +bx + 2c2x2 = a +bx +cx2

Examples: ln(1+x) ≅ x -x22, sinx ≅ x, cosx ≅ 1 -12x2 (1.d.), ex ≅ 1 + x + 12x2, (1+x)r = 1 +rx + r(r-1)2x2

The linear approximation of cosx near 0 is the straight horizontal line y = 1. This doesn’t seem like a very good approximation. The quadratic approximation to the graph of cos(x) is a parabola that opens downward, cosx ≅ 1 -12x2 (1.d.), this is much closer to the shape of its graph than the line y = 1.

Example. Find quadratic approximations for f(x)=xe-2x near x=1 and g(x)=e-3x(1+x)-1/2 near x=0.

f’(x)=e-2x-2xe-2x=e-2x(1-2x) ⇒ f’(1)= -e-2
f’’(x)=-2e-2x-2e-2x(1-2x)=-2e-2x(2-2x)=-4e-2x(1-x) ⇒ f’’(1)=0.

f(x) ≅ e-2 -e-2(x-1)

$e^{-3x}(1+x)^\frac{-1}{2} ≅ (1 -3x + \frac{9x^{2}}{2})(1 +\frac{-1}{2}x + \frac{3}{8}x^{2}) ≅ 1 -3x + \frac{9x^{2}}{2} +\frac{-1}{2}x + \frac{3}{2}x^{2} + \frac{3}{8}x^{2} ≅ 1 +\frac{-7}{2}x + \frac{51}{8}x^{2} $

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