Find derivative:
y= e^(x^x)
if y = u^v then
y' = vu^(v-1)u' + lnu u^v v'
Note that u^n and a^u are just special cases of this
So, what you have is
y= e^u
y' = e^u u'
u = x^x, so u' = x^x (1+lnx)
y' = e^(x^x) x^x (1+lnx)
let z = x^x
y = e^z
dy/dz = z e^z = x^x (e^x^x)
dy/dx = dy/dz * dz/dx so we need dz/dx when z = x^x
look here
http://www.analyzemath.com/calculus/Differentiation/first_derivative.html for
dz/dx = (ln x + 1) x^x
so
x^x ( e^x^x) (ln x + 1) x^x
=x^2x e^x^x (ln x + 1)
where did you get
d/dz e^z = z e^z?
To find the derivative of y = e^(x^x), we need to use the chain rule. The chain rule states that if we have a composite function, where we take the exponential of another function, we need to multiply the derivative of the outer function by the derivative of the inner function.
Let's break down the steps to find the derivative of y:
Step 1: Identify the outer and inner functions.
In our case, the outer function is e^u, where u = x^x. The inner function is u = x^x.
Step 2: Find the derivative of the inner function.
To find the derivative of u = x^x, we can use logarithmic differentiation or take the natural logarithm of both sides and then differentiate implicitly. Let's use the second method:
Take the natural logarithm of both sides:
ln(u) = ln(x^x)
Apply the power rule of logarithms:
ln(u) = x ln(x)
Implicitly differentiate both sides with respect to x:
(1/u) * du/dx = ln(x) + x(1/x)
Simplify the right side:
(1/u) * du/dx = ln(x) + 1
Multiply both sides by u:
du/dx = u * [ln(x) + 1]
Replace u with x^x:
du/dx = x^x * [ln(x) + 1]
Therefore, the derivative of u = x^x is du/dx = x^x * [ln(x) + 1].
Step 3: Find the derivative of the outer function.
The derivative of e^u = e^u.
Step 4: Apply the chain rule.
Now that we have the derivatives of both the outer and inner functions, we can apply the chain rule. The chain rule states that we multiply the derivative of the outer function by the derivative of the inner function.
dy/dx = (e^u) * (du/dx)
Substituting the expressions for e^u and du/dx:
dy/dx = (e^(x^x)) * [x^x * (ln(x) + 1)]
Therefore, the derivative of y = e^(x^x) is dy/dx = (e^(x^x)) * [x^x * (ln(x) + 1)].