<p>hey, I'm having trouble with easy problem, can someone tell me how to solve it?</p>
<p>d/dx ( (x^2 + 1 ) ^ ln x )</p>
<p>let y=(x^2+1)^ln(x)</p>
<p>take ln of both sides:
ln(y)= ln((x^2+1)^lnx)
= ln(x)*ln(x^2+1)</p>
<p>Take derivative of both sides:
y'/y= 1/x<em>ln(x^2+1) + ln(x)</em>(2x/(x^2+1))</p>
<p>Multiply both sides by y:</p>
<p>y'= y ( ln(x^2+1)/x + ln(x)*(2x/(x^2+1)))</p>
<p>Substitute for y=(x^2+1)^ln(x)
y'= ((x^2+1)^ln(x))*ln(x^2+1)/x + 2xln(x)/(x^2+1))</p>
<p>I haven't checked this, but I'm pretty sure it's right.</p>
<p>errr..you misplaced parentheses on the last step, but besides that, I got the same thing; y'= ln(x)<em>ln(x^2+1)</em> ( ln(x^2+1)/x + ln(x)*(2x/(x^2+1)))</p>