i want to find an exponential function?
95230
The compound interest formula is an exponential function:\n\nA=P(1+(r/n))^(n*t) where\nA=final amount\nP=principal or beginning amount\nr=interest rate (as a decimal)\nn=number of compounding per year\nt=time in years\n\nThe other exponential formula for compounding interest is used for continuously compounding interest or when (n*t) is equal to infinity. It is also the equation used when charting the growth or decay of bacteria.\n\nA=P*e^(r*t).\n\nPlease check the second source for info on how this equation relates to bacteria growth/decay.
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
1044811
Application 1 : Exponential Growth - Population\nLet P(t) be a quantity that increases with time t and the rate of increase is proportional to the same quatity P as follows\n\nd P / d t = k P\n\n\nwhere d p / d t is the first derivative of P, k > 0 and t is the time.\n\nThe solution to the above first order differential equation is given by\n\nP(t) = A ek t\n\n\nwhere A is a constant not equal to 0.\n\nIf P = P0 at t = 0, then\n\nP0 = A e0\n\nwhich gives A = P0\n\nThe final form of the solution is given by\n\nP(t) = P0 ek t\n\n\nAssuming P0 is positive and since k is positive, P(t) is an increasing exponential. d P / d t = k P is also called an exponential growth model.\n\nApplication 2 : Exponential Decay - Radioactive Material\nLet M(t) be the amount of a product that decreases with time t and the rate of decrease is proportional to the amount M as follows\n\nd M / d t = - k M\n\n\nwhere d M / d t is the first derivative of M, k > 0 and t is the time.\n\nSolve the above first order differential equation to obtain\n\nM(t) = A e- k t\n\n\nwhere A is non zero constant.\n\nIt we assume that M = M0 at t = 0, then\n\nM0 = A e0\n\nwhich gives A = M0\n\nThe solution may be written as follows\n\nM(t) = M0 e- k t\n\n\nAssuming M0 is positive and since k is positive, M(t) is an decreasing exponential. d M / d t = - k M is also called an exponential decay model.\n\nApplication 3 : Falling Object\nAn object is dropped from a height at time t = 0. If h(t) is the height of the object at time t, a(t) the acceleration and v(t) the velocity. The relationships between a, v and h are as follows:\n\na(t) = dv / dt , v(t) = dh / dt.\n\nFor a falling object, a(t) is constant and is equal to g = -9.8 m/s.\n\nCombining the above differential equations, we can easily deduce the follwoing equation\n\nd 2h / dt 2 = g\n\nIntegrate both sides of the above equation to obtain\n\ndh / dt = g t + v0\n\nIntegrate one more time to obtain\n\nh(t) = (1/2) g t + v0 t + h0\n\nThe above equation describes the height of a falling object, from an initial height h0 at an initial velocity v0, as a function of time.\n\nApplication 4 : Newton's Law of Cooling\nIt is a model that describes, mathematically, the change in temperature of an object in a given environment. The law states that the rate of change (in time) of the temperature is proportional to the difference between the temperature T of the object and the temperature Te of the environment surrounding the object.\n\nd T / d t = - k (T - Te)\n\n\nLet x = T - Te so that dx / dt = dT / dt\n\nUsing the above change of variable, the above differential equation becomes\n\nd x / d t = - k x\n\n\nThe solution to the above differential equation is given by\n\nx = A e - k t\n\nsubstitute x by T - Te\n\nT - Te = A e - k t\n\nAssume that at t = 0 the tempearture T = To\n\nTo - Te = A e 0\n\nwhich gives A = To - Te\n\nThe final expression for T(t) i given by\n\nT(t) = Te + (To - Te)e - k t\n\nThis last expression shows how the tempearture T of the object changes with time.\n\nApplication 5 : RL circuit\nrl circuit for application 5\nLet us consider the RL (resistor R and inductor L) circuit shown above. At t = 0 the switch is closed and current passes through the circuit. Electricty laws state that the voltage across a resistor of resistance R is equal to R i and the voltage across an inductor L is given by L di/dt (i is the current). Another law gives an equation relating all voltages in the above circuit as follows:\n\nL di/dt + Ri = E , where E is a constant voltage.\n\nLet us solve the above differential equation which may be written as follows\n\nL [ di / dt ] / [E - R i] = 1\n\nwhich may be written as\n\n- (L / R) [ - R d i ] / [E - Ri] = dt\n\nIntegrate both sides\n\n- (L / R) ln(E - R i) = t + c , c constant of integration.\n\nFind constant c by setting i = 0 at t = 0 (when switch is closed) which gives\n\nc = (-L / R) ln(E)\n\nSubstitute c in the solution\n\n- (L / R) ln(E - R i) = t + (-L/R) ln (E)\n\nwhic
164074
alright, finally a question i can answer, ok first of all u need some bivariate data (meaning data in the first two lists). exponential regression is going to find the equation of the line that best fits the data. on my calc there is linear regression (straight line), quadratic regression (parabola looking thing), cubic regression (cubic looking thing), quartic regression (weird looking parabola), and ooo i see what ur talking about. hmm i think its like finding an equation of the line then looks like an exponential equation. and example of this is graphing y=2^x. try that and u'll see what i mean
713982
use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
158690
substitute x=e^y\nthen use a table of integration formulas?
139157
Try to fit your data with Excel’s exponential fitting curves,\nIt will give you Coefficients which make the exponential looks like a line.\n\nOr try the experiment more rapidly.
797160
Your function y=4*x^2*e^(5x) is defined for any x. Meanwhile x^2>=0 & e^(5x)>0, thus y>=0 for any x. Now y’=4*(2x+5x^2)*e^(5x), local extrema being at y’=0. Or x1=0, x2=-0.4\nHence and “above” x1=0 means local min & y(0)=0, so x2=-0.4 means local max, \nLim[for x->-infinity]=0 & Lim[for x->+infinity]=+infinity. Does it help you Xu?
1130920
You can't solve this using traditional techniques since you'll either have a logarithm or an exponential on one of the x variables.  The people claiming Taylor series and techniques like Newton-Raphson approximation are most likely on the right track.  I would use a CAD program, or graph the function y(x)=(1+x)^2-2^x and find the zeroes that way by zooming and such (on at TI-83 or similar).\n\nIn this case, I'll use MathCAD.  The solutions are:\nx={-1.57862, 0, 5.31972} (two of these are approximate, by the way)
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
766600
You have to use the quotient rule for this problem:\n\nFor any function f(x) defined by, f(x) = g(x)/h(x),\n\nf'(x) = [g'(x)h(x) - h'(x)g(x)]/[h^2(x)]\n\nIn this problem, we have:\n\ng(x) = xe^-x\nh(x) = x^3+x\n\nTherefore, \n\ng'(x) = xe^-x(-1) + e^-x(1) [you have to use product and chain rule for thie derivative]\n\nSimplfying, you get e^-x(1-x)\n\nh'(x) = 3x^2 + 1 (power rule)\n\nNow that we have all parts of the derivative, just plug in:\n\nf'(x) = [e^-x(1-x)(x^3+x) - (3x^2+1)(xe^-x)]/(x^3+x)^2\n\nNow just simplify this ugly mess =P\n\n---------\n\nHope this helps
953689
A clue:\n\nsin h x = (e^x + e^-x) / 2\n\nCalculate e^(2-i) = e^2 e^-i, etc\n\nsin h x = (e^2 e^-i +e^-2 e^î)^2\n\nAna
26116
Exponential functions are used in the everyday lives of people who model (represent mathematically):\n\na) exponential growth - such as the change in a population over time.\n\nb) exponential decay - such as the change in radioactivity of a substance over time.
935326
You have got to use expansion series!\n\nX->0+ [e-(1+x)^(1/x)]/x =e/2 \nlet x=0+h\n=>as x->0, h->o\n\nh->0 [e-(1+h)^(1/h)]/h\n\nlet y=(1+h)^(1/h)\n=>log y=(1/h)log(1+h)\n=>y=e^[(1/h)log(1+h)]\n\nexpansion of log(1+x)=x - (x^2)/2..................\n=>e^[(1/h)log(1+h)]= e^[(1/h)[h - (h^2)/2!............]\n\n=>e^[(1/h)[-h - (h^2)/2!............]=e^[1-h/2!........]\n\ntherefore,\n\nlet y =lim h->0 [e-(1+h)^(1/h)]/h=lim h->0 {[e- e^(1-h/2!...)]/h}\n\nexpansion of e^x=1+x+(x^2)/2!.................\n\n=>lim h->0 {[e- e^(1-h/2...)]/h}=lim h->0 {-e[e^(-h/2.....) -1]}/h\n\nexpansion of e^x=1+x+(x^2)/2!.................\n=>lim h->0{-e[1+(-h/2-(h^2)/3.....)+ ........-1]/h}\n\n=>lim h->0{-e[(-h/2-(h^2)/2...)+ [(-h/2-(h^2)/2...)^2]/2!+.....]}/h\n\n=>lim h->0{-e(-1/2 -..(terms containing highter powers of h))}\n\n=>e/2\n\n\nP.S.  By the way L' Hospitals rule would fail here as limit is not of form 0/0 or ~/~. It is of form a/0 where a is constant.
738843
I'd recommend graphing the function on a calculator (I use a TI-83) and use the 'trace' to find zeros.
238752
ln and e^x are inverse functions. So you can use them to solve equations.\n\nA looks like it's already solved, so just use a calculator and find the ln of 0.0093, which is -4.67774\n\nFor equation B, take the ln of both sides.\n\nln(e^x) = ln(62.5)\n\nThe ln(e^x) cancels each other out.\n\nx = ln(62.5)\n\nx = 4.13517
119382
y=log_4 X\nand\ny= -log_3 (X+s) + 2\n\nRemember that logarithmic function y = log_b x is the inverse of the exponential function y = b^x, so you just graph its inverse (the exp function) and the graph of your log function is its mirror image with respect to the line \ny = x (The line that passes through (0,0) and has a slope of 1)\n\nIn your first given,\ny = log_4 x\nIts inverse function is\ny = 4^x\n\nNow graph y = 4^x\n\nNext, graph y = x and use this Line as your guide mirror to draw the log graph. This means that the shape of the log graph is the exact mirror image of the exp graph.\n\nIn your second given\ny= -log_3 (x + s) + 2\nJust rearrange the terms\ny - 2 = -log_3 (x + s)\n-(y - 2) = log_3 (x + s)\n\nThe inverse is\n-(y - 2) = 3^(x + s)\nGraph this one, then use y = x again as your "mirror."  Simple.\n\nMay this work of mine help you in your studies ;-)