f(x) = (x^2)(e^x)
f'(x) = e^x((x^2)+2x) - i think
f"(x) = ?
The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)
The derivative of ln x, the natural logarithm, is 1/x.Otherwise, given the identity logbx = log(x)/log(b), we know that the derivative of logbx = 1/(x*log b).ProofThe derivative of ln x follows quickly once we know that the derivative of ex is itself. Let y = ln x (we're interested in knowing dy/dx)Then ey = xDifferentiate both sides to get ey dy/dx = 1Substitute ey = x to get x dy/dx = 1, or dy/dx = 1/x.Differentiation of log (base 10) xlog (base 10) x= log (base e) x * log (base 10) ed/dx [ log (base 10) x ]= d/dx [ log (base e) x * log (base 10) e ]= [log(base 10) e] / x= 1 / x ln(10)
e^x/1-e^x
It is possible.
When the logarithm is taken of any number to a power the result is that power times the log of the number; so taking logs of both sides gives: e^x = 2 → log(e^x) = log 2 → x log e = log 2 Dividing both sides by log e gives: x = (log 2)/(log e) The value of the logarithm of the base when taken to that base is 1. The logarithms can be taken to any base you like, however, if the base is e (natural logs, written as ln), then ln e = 1 which gives x = (ln 2)/1 = ln 2 This is in fact the definition of a logarithm: the logarithm to a specific base of a number is the power of the base which equals that number. In this case ln 2 is the number x such that e^x = 2. ---------------------------------------------------- This also means that you can calculate logs to any base if you can find logs to a specific base: log (b^x) = y → x log b = log y → x = (log y)/(log b) In other words, the log of a number to a given base, is the log of that number using any [second] base you like divided by the log of the base to the same [second] base. eg log₂ 8 = ln 8 / ln 2 = 2.7094... / 0.6931... = 3 since log₂ 8 = 3 it means 2³ = 8 (which is true).
The first derivative of e to the x power is e to the power of x.
The derivative of ex is ex
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
y"+y'=0 is a differential equation and mean the first derivative plus the second derivative =0.Look at e-x the first derivative is -e-xThe second derivative will be e-xThe sum will be 0
I assume you mean 27 times e to the power x. 1) You take out the constant out. So, the derivative is 27 times the derivative of (e to the power x).2) You use the rule for the exponential function.
d/dx (e-x) = -e-x
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1: y = x y' = 1
Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x
ee is a constant and so its derivative is 0.
in case of derivative w.r.t time first derivative with a variable x gives velocity second derivative gives acceleration thid derivative gives jerk