can you help me differentiate y = x[sup:2jvdcgug](x^x)[/sup:2jvdcgug] thank you
A alexa_xox New member Joined Jan 23, 2009 Messages 7 Jan 28, 2009 #1 can you help me differentiate y = x[sup:2jvdcgug](x^x)[/sup:2jvdcgug] thank you
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Jan 28, 2009 #2 Re: dy/dx x^x^x Use logarithmic differentiation. \(\displaystyle y=x^{x^{x}}\) \(\displaystyle ln(y)=ln(x^{x^{x}})=x^{x}ln(x)\) \(\displaystyle \frac{y'}{y}=\underbrace{x^{x-1}\left(x(ln(x))^{2}+xln(x)+1\right)}_{\text{product rule}}\) Continue?. Remember what y equals when you solve for y'.
Re: dy/dx x^x^x Use logarithmic differentiation. \(\displaystyle y=x^{x^{x}}\) \(\displaystyle ln(y)=ln(x^{x^{x}})=x^{x}ln(x)\) \(\displaystyle \frac{y'}{y}=\underbrace{x^{x-1}\left(x(ln(x))^{2}+xln(x)+1\right)}_{\text{product rule}}\) Continue?. Remember what y equals when you solve for y'.
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Jan 30, 2009 #3 Re: dy/dx x^x^x I was bored, so here's a "neat" formula using the above method galactus lined out (that may be well known): \(\displaystyle \frac{dy}{dx}x^{f(x)} = x^{f(x)}f'(x)ln(x) + x^{f(x)-1}f(x)\) And a simpler formula for \(\displaystyle f(x)=x^n\) \(\displaystyle \frac{dy}{dx}x^{x^n} = x^{x^n+n-1}(nln(x)+1)\)
Re: dy/dx x^x^x I was bored, so here's a "neat" formula using the above method galactus lined out (that may be well known): \(\displaystyle \frac{dy}{dx}x^{f(x)} = x^{f(x)}f'(x)ln(x) + x^{f(x)-1}f(x)\) And a simpler formula for \(\displaystyle f(x)=x^n\) \(\displaystyle \frac{dy}{dx}x^{x^n} = x^{x^n+n-1}(nln(x)+1)\)
