Use the "chain rule" of differentiation: y=exp(exp(x)) taking ln both side in y=e x (1/y)dy/dx=e x dy/dx=y*e x dy/dx=exp(x+exp(x))
(1/2)x = 2-x = exp (ln 2-x) = exp( -x ln 2). Since d/dx exp(x) = exp(x), we can use the chain rule to find that: d/dx (1/2)x = -(ln 2) exp(-x ln 2).
(1/2)x = 2-x = exp (ln 2-x) = exp( -x ln 2). Since d/dx exp(x) = exp(x), we can use the chain rule to find that: d/dx (1/2)x = -(ln 2) exp(-x ln 2).
x e^x +C
Negative square roots are just the opposite of positive square roots. Since square roots (of positive numbers) are real, the negative square roots are also real.Square roots of negative numbers are not real.Note that -1 = exp(Pi*i), so (-1)^(1/2) = exp((1/2)*Pi*i) = i.Note that exp(i*x) = cos(x) + i*sin(x), for instance by taking derivatives:(d/dx)(exp(i*x)) = i*exp(i*x), and(d/dx)^2(exp(i*x)) =(-1)*exp(i*x).This means that the second derivative of exp(i*x) equals -exp(i*x).The same property holds for cos(x) + i*sin(x):(d/dx)(cos(x) + i*sin(x)) = -sin(x) + i*cos(x)(d/dx)^2(cos(x) + i*sin(x)) = -cos(x) - i*sin(x) = -(cos(x) + i*sin(x)))Hence cos(x) + i*sin(x)) = C + Dx + exp(i*x), for some C and D.Comparing the values on both sides for x = 0, we find:1 = C+1, so C = 0 and for the first derivative:i = D + i, so D = 0.So cos(x) + i*sin(x)) = exp(i*x) for all x.by comparing x=0 for both functions and their first derivative. Since they coincide,
Use the "chain rule" of differentiation: y=exp(exp(x)) taking ln both side in y=e x (1/y)dy/dx=e x dy/dx=y*e x dy/dx=exp(x+exp(x))
exp 4In X? I must assume you mean 4InX or InX^4 Regardless, the derivitive of InX = 1/X
It is -exp (-x) + C.
tanh is the hyperbolic tangent and it is computed as sinh(x)/cosh(x) = [exp(x)-exp(-x)]/[exp(x)+exp(-x)] and there are other ways of computing it, including infinite series.
(1/2)x = 2-x = exp (ln 2-x) = exp( -x ln 2). Since d/dx exp(x) = exp(x), we can use the chain rule to find that: d/dx (1/2)x = -(ln 2) exp(-x ln 2).
(1/2)x = 2-x = exp (ln 2-x) = exp( -x ln 2). Since d/dx exp(x) = exp(x), we can use the chain rule to find that: d/dx (1/2)x = -(ln 2) exp(-x ln 2).
x e^x +C
Negative square roots are just the opposite of positive square roots. Since square roots (of positive numbers) are real, the negative square roots are also real.Square roots of negative numbers are not real.Note that -1 = exp(Pi*i), so (-1)^(1/2) = exp((1/2)*Pi*i) = i.Note that exp(i*x) = cos(x) + i*sin(x), for instance by taking derivatives:(d/dx)(exp(i*x)) = i*exp(i*x), and(d/dx)^2(exp(i*x)) =(-1)*exp(i*x).This means that the second derivative of exp(i*x) equals -exp(i*x).The same property holds for cos(x) + i*sin(x):(d/dx)(cos(x) + i*sin(x)) = -sin(x) + i*cos(x)(d/dx)^2(cos(x) + i*sin(x)) = -cos(x) - i*sin(x) = -(cos(x) + i*sin(x)))Hence cos(x) + i*sin(x)) = C + Dx + exp(i*x), for some C and D.Comparing the values on both sides for x = 0, we find:1 = C+1, so C = 0 and for the first derivative:i = D + i, so D = 0.So cos(x) + i*sin(x)) = exp(i*x) for all x.by comparing x=0 for both functions and their first derivative. Since they coincide,
exp(x)
cos x
The definite integral of the function f=x*exp(k*x) is (1/k)*(x-(1/k))*exp(k*x). So you have the answer to your questions by setting k equal to 1 then 2. I derived my formula by using integration by parts, setting u=x and dv=exp(k*x)dx.
3 sec23x