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. y=xxx\displaystyle y=x^{x^{x}}y=xxx ln(y)=ln(xxx)=xxln(x)\displaystyle ln(y)=ln(x^{x^{x}})=x^{x}ln(x)ln(y)=ln(xxx)=xxln(x) y′y=xx−1(x(ln(x))2+xln(x)+1)⏟product rule\displaystyle \frac{y'}{y}=\underbrace{x^{x-1}\left(x(ln(x))^{2}+xln(x)+1\right)}_{\text{product rule}}yy′=product rulexx−1(x(ln(x))2+xln(x)+1) Continue?. Remember what y equals when you solve for y'.
Re: dy/dx x^x^x Use logarithmic differentiation. y=xxx\displaystyle y=x^{x^{x}}y=xxx ln(y)=ln(xxx)=xxln(x)\displaystyle ln(y)=ln(x^{x^{x}})=x^{x}ln(x)ln(y)=ln(xxx)=xxln(x) y′y=xx−1(x(ln(x))2+xln(x)+1)⏟product rule\displaystyle \frac{y'}{y}=\underbrace{x^{x-1}\left(x(ln(x))^{2}+xln(x)+1\right)}_{\text{product rule}}yy′=product rulexx−1(x(ln(x))2+xln(x)+1) 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): dydxxf(x)=xf(x)f′(x)ln(x)+xf(x)−1f(x)\displaystyle \frac{dy}{dx}x^{f(x)} = x^{f(x)}f'(x)ln(x) + x^{f(x)-1}f(x)dxdyxf(x)=xf(x)f′(x)ln(x)+xf(x)−1f(x) And a simpler formula for f(x)=xn\displaystyle f(x)=x^nf(x)=xn dydxxxn=xxn+n−1(nln(x)+1)\displaystyle \frac{dy}{dx}x^{x^n} = x^{x^n+n-1}(nln(x)+1)dxdyxxn=xxn+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): dydxxf(x)=xf(x)f′(x)ln(x)+xf(x)−1f(x)\displaystyle \frac{dy}{dx}x^{f(x)} = x^{f(x)}f'(x)ln(x) + x^{f(x)-1}f(x)dxdyxf(x)=xf(x)f′(x)ln(x)+xf(x)−1f(x) And a simpler formula for f(x)=xn\displaystyle f(x)=x^nf(x)=xn dydxxxn=xxn+n−1(nln(x)+1)\displaystyle \frac{dy}{dx}x^{x^n} = x^{x^n+n-1}(nln(x)+1)dxdyxxn=xxn+n−1(nln(x)+1)
