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what is the formula for calculating the instantaneous rate of change for f(x)=e^x where x=2

I need to find the instantaneous rate of change for f(x) = e^x, x=2

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Finding the instantaneous rate of change of the function $f(x ...


Finding the instantaneous rate of change of the function $f(x)=-x^2+4x$ at $x=5$, I know the formula for instantaneous ... The instantaneous rate of change, i.e. the ...
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Instantaneous Rate of Change Formula | [email protected]


The Instantaneous Rate of Change Formula provided with limit exists is, ... So, the instantaneous rate of change at x = 2 f'(2) = 15(2) 2 - 8(2) + 2 = 60 - 16 + 2 = 46 .
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what is the formula for calculating the instantaneous rate of ...


I need to find the instantaneous rate of change for f(x) = e^x, x=2
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Average and Instantaneous Rates of Change - UUMath - Home


9.3 Average and Instantaneous Rates of Change: ... this instantaneous rate of change of a function f is ... of x, say by evaluating the derivative formula ...
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Instantaneous Rate of Change - Virginia Tech


A logical extrapolation would indicate that the instantaneous rate of change in F(x) ... 2 dy x dx = is the general slope formula for any tangent line touching f (x) ...
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5.1 Average and instantaneous rate of change


the instantaneous rate at x = 3. In fact that’s more or less how the general theory goes. Formally, the instantaneous rate of change of f(x) ...
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Finding Instantaneous Rate of Change of a Function: Formula ...


In this lesson, you will learn about the instantaneous rate of change of a function, or derivative, and how to find one using the concept of limits...
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Calculus Online Book


Free online Calculus e-book focusing on understanding ... we can define the instantaneous rate of change of a function ... from x = 0 to x = 2 the change in f ...
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Instantaneous Rate of Change - [email protected]


It should be noted that the instantaneous rate of change of $y = f(x) ... The formula for instantaneous rate of change is ... Calculating Instantaneous Rate of Change.
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Suggested Questions And Answer :


I need help for this problem. Thanks. I really appreciate the help. How do I find theta?

The minute hand moves 12 times as fast as the hour hand, because the hour hand moves 1/12 of the area of the clock face while the minute hand moves the whole area of the clock face. At 4 o'clock, the hands are 120 degrees apart (one third of 360 degrees). The tips of the hands at this time are distance x apart where x is given by the cosine rule: x=sqrt(3^2+4^2-2*3*4cos120)=sqrt(25+12)=sqrt(37), because cos120=-1/2. At time t minutes after 4, the angle between the tips of the hands decreases initially. For example, if t=15 minutes, the hour hand will move 1/4 the distance between 4 and 5, i.e., 30/4=7.5 degrees, while the minute will have moved 90 degrees. The angle between the hands becomes 120-90+7.5=37.5 degrees. For t minutes, then, the angle ("theta") changes to 120-360t/60+30t/60 degrees=120-6t+t/2=120-11t/2. (Theta can be found for any time t minutes after 4 o'clock from this formula. Also, t can be found when theta is zero, the hands being aligned: t=240/11.) So, x=sqrt(25-24cos(120-11t/2))=sqrt(25-24(cos120cos(11t/2)+sin120sin(11t/2)). If t is very small cos(11t/2) is close to 1 and sin(11t/2) is close to 11t/2. sqrt(120)=sqrt(3)/2; x=sqrt(25-24(-1/2+11tsqrt(3)/4))=sqrt(37-264tsqrt(3)/4)=sqrt(37-66tsqrt(3)). The change in x is sqrt(37-66tsqrt(3))-sqrt(37)=sqrt(37)(sqrt(1-66tsqrt(3)/37)-1). We can use the Binomial Theorem to evaluate sqrt(1-66tsqrt(3)/37)=(1-66tsqrt(3)/37)^(1/2). When t is very small we can ignore t^2 terms and higher powers of t, so we get 1-33tsqrt(3)/37 as an approximation. The change in x becomes sqrt(37)(1-33tsqrt(3)/37-1)=-33tsqrt(3/37). The instantaneous rate of change of x at 4 o'clock is found by dividing this expression by t=-33sqrt(3/37)=-9.3967"/minute=-0.1566"/sec. The minus sign indicates that the hands are closing at the rate of 0.1566 inches per second. The solution can also be found applying formal calculus, where dx (the change in x) is related to the angle change (dø) and the angular rate of change dø/dt can be related to the rate of change of x, dx/dt. This involves differentiating the cosine expression. Because a time of 4 o'clock has been specified, the solution has been calculated from first principles, thus avoiding calculus.  
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what is the formula for calculating the instantaneous rate of change for f(x)=e^x where x=2

d/dx av e^x=e^x...1 av the basik propertees av e e^x at x=2=e^2 or about 7.389046
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For the function f(x)=x^2-4x:

a. The rate of change is given by f'(x)=2x-4. The average rate of change if (f'(4)+f'(1))/2=(4-2)/2=1. 2x-4 is a linear function, and if we take integer values of x and average them we get: (-2+0+2+4)/4=1. If we increment x by 0.5 steps we get: (-2-1+0+1+2+3+4)/7=1, and so on. b. Instantaneous change of direction at x=1 is -2.
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area of a square with side length x is x2. Instantaneous rate of change in area if x is 5

area of a square with side length x is x^2. Instantaneous rate of change in area if x is 5 We can write the area A as, A = x^2 where the area, A, is a function of x. The rate of chabge of A is dA/dx, where dA/dx = 2x When x = 5, then dA/dx = 2*5 = 10 Answer: Rate of change of area of square, when side length x = 5, is 10
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A =(p+i)^n make i the subject of the fomular?? find i if A = 2 200 P=200 and n =5

A=(p+i)^n; p+i=A^(1/n) [A^(1/n) is the same as the nth root of A]; i=A^(1/n)-p. A=2,200, p=200, n=5: i=2200^0.2-200=-195.34 approx. As a percentage i is-19,534%. [IMPORTANT NOTE: This problem looks like a variation on the formula for calculating compound interest which is contained in two formulae: A=p(1+r/100)^n and i=A-p, making i=p((1+r/100)^n-1), where r=percentage interest rate for a period, p is the initial amount, A is the total amount after n periods of time, and i is the interest. If this what the question is all about then we can calculate r: A/p=(1+r/100)^n; (A/p)^(1/n)=1+r/100; r=100((A/p)^(1/n)-1); r=100((2200/200)^0.2-1)=61.54% (0.6154 as a decimal); i=A-p=2000, which is 10 times the amount invested. If the original formula had been A=p(1+i)^n, then i=r/100 and i is not the interest, but the interest rate expressed as a decimal rather than as a percentage, so i=0.6154 when r=61.54%. In words, the problem could be expressed: If RM200, invested at compound interest over 5 years, gains 10 times this amount in interest, what is the annual compound percentage interest rate? The answer would be 61.54%.]  
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How much would density be of each planet have to change to have same volume as Earth?

The general form for density D=M/V. Applying this formula to the Earth, DE=ME/VE. So, if V is the volume of a planet, V/VE is the proportion of its volume in relation to Earth. If the planet is bigger than Earth, then V/VE>1 and compressing its mass into the same volume as the Earth would increase its density by the same factor. So the density would change to D.V/VE. There would be a decrease in density if V Read More: ...

why is the rate of fall 32 feet per second squared instead of 16 feet per second squared since 16 feet is how far it fell in the first second ?

Objects fall under gravity at an accelerating rate of 32 feet per second per second. That means the speed is changing at the rate of (32 ft/sec) per second. An object which is simply dropped (not thrown) will pick up speed so that after the first second it will be travelling at 32 feet per second. After the second second its speed will have reached 64 feet per second. That's how the speed changes, but what about the distance? To calculate this we start with a speed of zero since we only just let the object fall. After one second the speed is 32 feet per second, so its average speed over the first second is half of the sum of the speed at the start and the speed after one second=1/2(0+32)=16 ft/sec. That's why in one second the object only travels 16 feet. How far does it travel in 2 seconds? This time the average speed is 1/2(0+64)=32 f/s so the distance travelled in 2 secs is 64 feet. How far did it travel between the first and second seconds? Average speed is 1/2(32+64)=48 feet per second. So between the 1st and 2nd second it travelled 48 feet. 48+16=64. The acceleration is constant but the speed varies proportionately. This gives rise to the equation s=1/2at^2 for every object falling under its own weight without being thrown, where a is the acceleration due to gravity and t=time. Put t=2 and a=32 in this equation and you can see it fits the calculations.
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Find the instantaneous rate of change of the curve f(x)=(x)/(2-3x) at x=2 using first principles


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What do I need to convert for this problem and how?

You need to convert time and speed into distance. Distance=(speed)*(time). Man1 starts at 1 pm and by 4 pm he has walked for 3 hours, and his distance is 3*3=9 miles. Man2 starts at 2 pm and by 4pm he has walked for 2 hours, and his distance is 2*4=8 miles. So we have a triangle and the included angle of 60 degrees. At 4 pm they are separated by sqrt(9^2+8^2-2*8*9cos60)=sqrt(81+64-144/2)=sqrt(145-72)=sqrt(73)=8.544 miles (cosine rule, and  cos60=1/2). To work out how fast they are separating we need to work out the rate of change of the distance between them. Let's suppose that they each walk for another h hours, where h is a very small fraction. Man1 walks a further 3h miles and Man2 a further 4h miles. Their separation is again given by the cosine rule: sqrt((9+3h)^2+(8+4h)^2-2(9+3h)(8+4h)cos60)= sqrt(81+54h+9h^2+64+64h+16h^2-72-60h-12h^2)= sqrt(73+58h+13h^2). [As an example, put h=1 hr, which is not a small value. The separation is then sqrt(144)=12. So the rate of change in separation is 12-sqrt(73)=3.456 mph.] This means that their separation has increased by sqrt(73+58h+13h^2)-sqrt(73). Because h is very small we can ignore h^2 terms and just consider sqrt(73+58h)-sqrt(73)=sqrt(73)*sqrt(1+58h/73)-sqrt(73) or sqrt(73)(sqrt(1+58h/73)-1). We can now use the Binomial Theorem to expand (1+58h/73)^(1/2)=1+29h/73 ignoring terms with h^2 and beyond as insignificant. So now we have: sqrt(73)(1+29h/73-1)=29hsqrt(73)/73. This is the separation just after 4 pm, h hours after 4 pm in fact. So the speed of separation is this expression divided by the time, h=29sqrt(73)/73 or 29/sqrt(73)=3.3942 mph. [Compare this to the value when h=1 hr, when the average rate of change was 3.456 mph.] What we have just done is calculated the results that calculus would have given us if we had formally applied it. By taking h as very small, the result we obtained applies to a vanishingly small h, or h approaches zero.  
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Find the average rate of change of y with respect to x in the interval from x=2 to x=3

The question doesn't give y in terms of x so this answer is general. Calculate dy/dx and call the result g(x). The average rate of change is (g(2)+g(3))/2.
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