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How do you differentiate exp exp exp x?
By using the chain rule. Since the derivative of exp(x) is exp(x), the derivative of exp(exp(exp(x))) is exp(exp(exp(x))) times the derivative of what is inside the parentheses, i.e., exp(exp(exp(x))) times derivate of exp(exp(x)). Continue using the chain rule once more, for this expression.
What is the derivative of a half to the power of 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).
What is the derivative of half to the power of 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).
What is the derivative of -exp to the -x?
x e^x +C
Why are negative square roots are on the real number line if square root of a negative number not a real number?
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,
How do you differentiate exp exp exp x?
By using the chain rule. Since the derivative of exp(x) is exp(x), the derivative of exp(exp(exp(x))) is exp(exp(exp(x))) times the derivative of what is inside the parentheses, i.e., exp(exp(exp(x))) times derivate of exp(exp(x)). Continue using the chain rule once more, for this expression.
How do you differentiate exp 4ln x?
exp 4In X? I must assume you mean 4InX or InX^4 Regardless, the derivitive of InX = 1/X
What is the antiderivative of exp -x?
It is -exp (-x) + C.
What is tanh?
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.
What is the derivative of a half to the power of 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).
What is the derivative of half to the power of 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).
What is the derivative of -exp to the -x?
x e^x +C
Why are negative square roots are on the real number line if square root of a negative number not a real number?
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,
How do you write exponential in scilab?
exp(x)
Differentiate sin x with respect to x?
cos x
How is xex and xe2x integrated manually?
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.
Differentiate tan3 x?
3 sec23x
