Question

Can someone tell me where my error is in finding the derivative of `y=x^(lnx)`?

I'm using the rule `d/dxa^u = a^u ln a (du)/dx`.

`dy/dx = x^(lnx) * ln x * 1/x`

`=(x^(lnx) ln x)/x`

The correct answer is `=(2x^(lnx) ln x)/x`.

Answer 1

`dy/dx = (2x^lnxlnx)/x`

Explanation:

I'm pretty fond of logarithmic differentiation. We try to take the derivative of both sides:

`lny = ln(x^(lnx))`

`lny = lnxlnx`

Now use implicit differentiation and the product rule.

`1/y(dy/dx) = 1/xlnx + 1/xlnx`

`1/y(dy/dx) = 2/xlnx`

`dy/dx = y(2/xlnx)`

`dy/dx= x^lnx(2/xlnx)`

Or

`dy/dx = (2x^lnxlnx)/x`

As required.

Hopefully this helps!

Answer 2

I got `dy/dx=(2ln(x)x^lnx)/x` which can be rewritten as `dy/dx=(2x^lnxln(x))/x`

Explanation:

You've used the wrong derivative technique!

`d/dxa^u=a^u lna (du)/dx`

This can only be used when `a` is a constant such as:

`5^x->d/dx=5^xln(5)`

But we have an `x` as the base!

So we must use a special technique which involves talking the natural log `(ln)` of both sides and using implicit differentiation

Given: `y=x^lnx`

Take `ln` of both sides

`ln(y)=ln(x^lnx)`

Since `color(blue)(ln(x^a)=aln(x)`

`ln(y)=ln(x)*ln(x)`

Differentiate both sides W.R.T `x` (The right side requires the product rule)

For the left side: `d/dx(ln(y))=dy/dx*1/y`

For the right side:

Product rule: `d/dx(f(x)*g(x))=f'(x)g(x)+f(x)g'(x)`

Let `f(x)=ln(x)` and `g(x)=ln(x)`

Thus `f'(x)=1/x` and `g'(x)=1/x`

`dy/dx*1/y=1/x*ln(x)+ln(x)*1/x`

`dy/dx*1/y=ln(x)/x+ln(x)/x`

`dy/dx*1/y=(2ln(x))/x`

Multiply both sides by `y`

`dy/dx=(2ln(x))/x*color(red)(y`

We want to rewrite everything in terms of `x` but we have this `color(red)(y` in the way. However, recall that `color(red)(y)` is defined as `color(red)(y=x^lnx)`

`dy/dx=(2ln(x))/x*color(red)(x^lnx`

Rewriting we get:

`dy/dx=(2ln(x)x^lnx)/x`

`color(red)("This can also be rewritten as"` `color(red)(dy/dx=(2x^lnxln(x))/x` `color(red)("which correlates to the correct answer you provided."`