How do you differentiate #f(x)=(ln(x))^x #?

Question

1569 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-12T00:41:53

Take the natural log of both sides, then use implicit differentiation ...

Take natural log of both sides, then use the property of logs :

#lny=ln(lnx)^x=xln(lnx)#

Now, using implicit differentiation, product and chain rules ...

#(1/y)y'=ln(lnx)+(x/lnx)xx1/x=ln(lnx)+1/lnx#

Finally, solve for #y'#

#y'=y[ln(lnx)+1/lnx]=(lnx)^x[ln(lnx)+1/lnx]#

hope that helped

Answer 2 · 2015-12-12T03:52:18

(IGNORE)

Rewrite #f(x)# using the properties of logarithms.

#f(x)=xln(x)#

Now, to find #f'(x)#, use the product rule.

#f'(x)=ln(x)d/dx[x]+xd/dx[ln(x)]#

Find each derivative.

#d/dx[x]=1#

#d/dx[ln(x)]=1/x#

Plug the derivatives back in.

#f'(x)=1(ln(x))+x(1/x)#

#f'(x)=ln(x)+1#