# How do you differentiate #(x(lnx)^(x-1))(1/x)#?

I would first rewrite the function:

Now use logarithmic differentiation:

By signing up, you agree to our Terms of Service and Privacy Policy

To differentiate the given expression (x(\ln(x))^{x-1}\cdot \frac{1}{x}), you can use the product rule and the chain rule.

Let's break it down:

- Product Rule: ((fg)' = f'g + fg')
- Chain Rule: ((g(h(x)))' = g'(h(x)) \cdot h'(x))

Now, let (f(x) = x) and (g(x) = (\ln(x))^{x-1}). Then, applying the product rule:

[f'(x) = 1] [g'(x) = (x-1)(\ln(x))^{x-2} \cdot \frac{1}{x}]

Now, applying the chain rule:

[g'(x) = (x-1)e^{(x-2)\ln(\ln(x))} \cdot \frac{1}{x}]

Now, let's put it all together:

[f'g = 1\cdot (\ln(x))^{x-1}] [fg' = x(\ln(x))^{x-1} \cdot (x-1)e^{(x-2)\ln(\ln(x))} \cdot \frac{1}{x}]

Therefore, the derivative of the given expression is:

[x(\ln(x))^{x-1} \cdot \frac{1}{x} + x(\ln(x))^{x-1} \cdot (x-1)e^{(x-2)\ln(\ln(x))} \cdot \frac{1}{x}]

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7