Question 541814
We know that {{{(d/dx)ln(x)=(1/x)dx}}}
So use a plug. Instead of {{{2x^5}}}, use "u"
{{{(d/du)ln(u)=(1/u)(du)}}}
{{{u=2x^5}}}
{{{du=10x^(4)dx}}}
So {{{(d/du)ln(u)=(1/u)(du)=(1/(2x^5))(10x^(4)dx)=(10x^4/2x^5)*dx=(5/x)dx}}}
Therefore, {{{(dy/dx)ln(2x^5)=(5/x)dx}}}
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That shows you how's it's done. The super easy way from here on out, just learn that {{{(d/dx)ln(c*x^n)=(n/x)dx}}}
"C" = any constant, and is irrelevant. Just put the exponent over x and you're done! :-)
Hope this helps!