exponential and logarithmic differentiation?
920193
e^(xlnx )=x^x\nderivative is  1+ln x
316594
The anti-derivative of x^(2)*e^(x)^(3) is (1/3)*e^(x)^(3)\n\nYou can always check by taking the derivative of your answer.
347784
Taking natural logarithm to both sides\nLn(y)= Ln(x) * Ln(x) = (Ln(x))^2\nDifferentiating\n(1/y )dy/dx = (2/x) * (Ln(x))\ndy/dx = (2y/x) * (Ln(x))\nat the point (e,e),\ndy/dx = (2e /e) * Ln(e)\nso the slope = 2
533385
Please believe TheoDorkx. He's the only one who understands logarithms.\n\nAnother way to solve logarithm problems is directly change from the logarithmic to the exponential form.\n\nln(2x) = 2 means that 2 is the power to which you raise the base of the logarithm, e, in order to get the result 2x.\n\nSo the equivalent exponential form is\n\n2x=e^2\n\ntherefore\n\nx=(1/2)e^2\n\nThis is the same thing TheoDorkx got with his approach. I offer it as a slightly different way to think about the problem.
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
756495
log(base x) y=[log(base 10)y]/[log(base10)x]. I hope you can take up from here!
940999
Differentiate the function as you would normally except that when the derivative of y is taken make sure to replace y with y'. So this would be,\n(1/2)(xy)^(-2/3)(xy') + y = 0 - y'\nThen, solve for y' by moving all the y' over to one side and factoring it out.\ny' + (1/2)(xy)^(-2/3)(xy') + y = 0\n\ny'((1/2)(xy)^(-2/3)x + 1) = -y\n\ny' = -y / ((1/2(xy)^(-2/3)x + 1)\n\nThe second derivative is (quotient rule),\ny'' = (.5(xy)^(-2/3) + 1)(-y') - (-y)(1/3)(xy)^(-5/3)xy(xy')(y)\n\ny'' = .5(xy)^(-2/3)y' - y' + (1/3)x^2*y^3*y'(xy)^(-5/3)\nFactor,\ny'' = y'{(xy)^(-5/3)[.5 - 1 + (1/3)x^2*y^3(xy)]}\n\ny'' = y'{(xy)^(-5/3)[- .5 + (1/3)x^2*y^3(xy)]}\n\n(check my work)
496208
log(x) - log(1 + (x^2 + 1)^(1/2))\n\nWhere log is the natural logarithm, of course.
1052815
3.1 - Given the equation of an exponential function, state its domain and graph it. \n\n3.2 - (a) Given a logarithmic equation, express it as an exponential equation.\n(b) Given an exponential equation, express it as a logarithmic equation. \n\n3.3 - Given the equation of a logarithmic function, state its domain and graph it. \n\n3.4 - (a) State the product, quotient and power properties for logarithmic functions.\n(b) Apply the properties of logarithms to expand and/or simplify expressions involving logarithms. \n\nSolving Exponential and Logarithmic Equations\n- Using the laws of exponents and the properties of logarithms, simplify expressions involving the composition of exponential and logarithmic function of the same base. \n - Given an equation involving logarithmic functions, solve it. \n\n - Given an equation involving exponential functions, solve it. \n\n- Given an applied problem described by an exponential or logarithmic equation, solve the
865054
Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.
108610
If look to the integrator more deeply, one finds that 1/x is the derivative of ln(x).\nThen this question comes up: \nWhat is the integration of uu'dx?\n\nObviously the answer is u^2/2.\n\nSo our answer is (ln(x))^2/4+c.
643522
The derivative of e^[f(x)] is f'(x)e^[f(x)]\nThat is, it is the derivative of the exponent times e raised to the original exponent.\ny=3xe^(5x)\nYou have to use the product rule.\ny' = (3x)'e^(5x) + (3x)(e^(5x))\n= 3e^(5x) + (3x)(5)(3^(5x))\n= 3e^(5x) + 15xe^(5x)\n\ny=(9x+5)^(3/5)+e^(3/5)\ny' = (3/5)((9x+5)^(-2/5))(9) + 0\n= (27/5)(9x+5)^(-2/5)\nsince there was no x involved in e^(3/5), then e^(3/5) is just a constant (and the derivative of a constant is 0)\nRemember: When taking the derivative of e, the exponent NEVER changes.
303836
Nope,\n\n   The log function blows up to infinity.  (If you don't believe this, give me a value for y=f(x) = K * log(x).  Ha ha! I can give you a value for x such that f(x) is greater than the value you gave me! )\n\n  Your pal from Calc I\n   cuntsucker
1090386
What?\nIf u=x^x\ndu/u=(1+ln(x))dx\n\nInteral  exp(x^x)dx=integral exp(u)du\n\nAnalytical solution for this indefinite integral is dificult.
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.
1154343
Use these facts: ln(x) is the natural logarithm of x and so\nln(e^x) = x and d/dx ln(x) = 1/x\n\nTo prove d/dx e^x = e^x you find the derivative in two ways, and equate them:\n\nFirst, trivially, since ln(e^x) = x it follows that:\n\nd/dx ln(e^x) = dx/dx = 1\n\nBut also, if we let u=e^x,  then by the chain rule:\n\nd/dx ln(e^x) = d/dx ln(u) = d/du log(u)*du/dx = \n\n1/u du/dx = 1/e^x d/dx e^x (since u = e^x)\n\nEquating these 2 results we have:\n\n1/e^x d/dx e^x = 1  and multiplying through by e^x we have\n\nd/dx e^x = e^x\n\nHope that's clear enough. You are doing very well for your age. Keep up the good work!
729730
u cant split a log like that! ie ln(ab) = lna +lnb\n\nu should use the f'(x) f(x)^n = f(x)^(n+1)/ (n+1)