How do I find the derivative of
x^ (1/(x-1))
is it
(1/(x-1)) x^[1-(1/(x-1))] * (-1/ (x-1)^2)
??
If not, can you please write out what rule I should use?
Thank you very much!
Use the chain rule. Let u = 1/(x-1)
u-1 = (1-x +1)/x-1 = -x/(x-1)
F(x) = x^u
dF/dx = dF/du * du/dx
= u*x^(u-1) * [-1/(x-1)^2]
= [1/(x-1)]*x^[-x/(x-1)]*[-1/(x-1)^2]
Our middle terms seem to differ.
y = x^(1/(x-1))
take ln of both sides
ln y = 1/(x-1) ( lnx)
lny = (x-1)^-1 (lnx)
now take derivative
(dy/dx)/y = -1(x-1)^-2 (lnx) + (x-1)^-1 (1/x)
dy/dx = y[-1(x-1)^-2 (lnx) + (x-1)^-1 (1/x)]
or
dy/dx = x^(1/(x-1))[-1(x-1)^-2 (lnx) + (x-1)^-1 (1/x)]
To find the derivative of the function f(x) = x^(1/(x-1)), you can use the chain rule.
Let's first take the natural logarithm of both sides of the equation: ln(f(x)) = ln(x^(1/(x-1))).
Next, we can apply the logarithm rule to simplify the expression: ln(f(x)) = (1/(x-1)) * ln(x).
Now, we can differentiate both sides of the equation with respect to x using the chain rule. The chain rule states that if y = f(u) and u = g(x), then dy/dx = f'(u) * g'(x).
So, differentiating both sides of the equation ln(f(x)) = (1/(x-1)) * ln(x) with respect to x, we get:
(d/dx)[ln(f(x))] = (d/dx)[(1/(x-1))*ln(x)].
On the left side of the equation, the derivative of ln(f(x)) is 1/f(x) * f'(x), according to the chain rule. On the right side, we need to apply the product rule and chain rule.
The product rule states that if u = f(x) * g(x), then d(u)/dx = f'(x) * g(x) + f(x) * g'(x).
Applying the product rule and chain rule to the right side of the equation, we have:
1/f(x) * f'(x) = (1/(x-1)) * ln(x)' + ln(x) * ((1/(x-1))' * (1/(x-1))),
Simplifying this expression, we get:
f'(x)/f(x) = (1/(x-1)) * (1/x) + ln(x) * (-1/(x-1)^2).
Multiplying both sides by f(x) and rearranging, we have:
f'(x) = f(x) * [(1/(x-1)) * (1/x) + ln(x) * (-1/(x-1)^2)].
Finally, substituting f(x) back in as x^(1/(x-1)), we have:
f'(x) = x^(1/(x-1)) * [(1/(x-1)) * (1/x) + ln(x) * (-1/(x-1)^2)].
Therefore, the derivative of x^(1/(x-1)) is (1/(x-1)) * (1/x) + ln(x) * (-1/(x-1)^2).