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#.

2 Answers
Jan 16, 2018

#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!

Jan 16, 2018

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."#