y=sqrt(x)^1/x. Find dy/dx
That is to the power of 1/x
So to start this do I need to take Ln on both sides and then use a logarithm property?
Ln(y)=lnx^1/2x?
that's the way it is usually done.
lny = 1/x (1/2 lnx)
1/y y' = -1/x^2 (1/2 lnx) + 1/x (1/2x)
= -1/(2x^2) (lnx + 1)
y' = -sqrt(x)^(1/x) / (2x^2) (lnx+1)
or
-1/2 x^(1/(2x)-2) (lnx+1)
Using the same method, you can work out that if
y = u^v
y' = v u^(v-1) u' + lnu u^v v'
You can see that it a combination/generalization of the derivatives of
u^n and a^u for constants a and n.
To find the derivative of y with respect to x, you can use logarithmic differentiation. Here's how you can proceed:
Step 1: Take the natural logarithm (ln) on both sides of the equation to simplify the expression:
ln(y) = ln(sqrt(x)^(1/x))
Step 2: Use the logarithmic identity for a power function to simplify the expression further:
ln(y) = (1/x) ln(sqrt(x))
Step 3: Apply the logarithmic differentiation property, which states that if y = f(x) and y = g(x)^h(x), then dy/dx = y * (h'(x) * ln(g(x)) + (g'(x) / g(x)) * h(x)).
In our case, let f(x) = y, g(x) = sqrt(x), and h(x) = 1/x. Applying the logarithmic differentiation property, we have:
dy/dx = y * (h'(x) * ln(g(x)) + (g'(x) / g(x)) * h(x))
Step 4: Calculate the derivatives of g(x) and h(x):
g'(x) = (1/2) * (x)^(-1/2) = 1 / (2 * sqrt(x))
h'(x) = -1/x^2
Step 5: Substitute the derivatives back into the previous equation:
dy/dx = y * (-1/x^2 * ln(sqrt(x)) + (1 / (2 * sqrt(x))) * (1/x))
Step 6: Substitute y = sqrt(x)^(1/x) back into the equation:
dy/dx = sqrt(x)^(1/x) * (-1/x^2 * ln(sqrt(x)) + (1 / (2 * sqrt(x))) * (1/x))
So, the derivative of y with respect to x, dy/dx, is given by the expression above.