Jan 10, 2004 #2 master_coda Messages 591 Reaction score 0 \begin{align*}<br /> y&=x^y \\<br /> \ln y&=y\ln x \\<br /> \frac{y^\prime}{y}&=y^\prime\ln x+\frac{y}{x} \\<br /> y^\prime\left(\frac{1}{y}-\ln x\right)&=\frac{y}{x} \\<br /> y^\prime&=\frac{y^2}{x-xy\ln x}<br /> \end{align*}
\begin{align*}<br /> y&=x^y \\<br /> \ln y&=y\ln x \\<br /> \frac{y^\prime}{y}&=y^\prime\ln x+\frac{y}{x} \\<br /> y^\prime\left(\frac{1}{y}-\ln x\right)&=\frac{y}{x} \\<br /> y^\prime&=\frac{y^2}{x-xy\ln x}<br /> \end{align*}
Jan 10, 2004 #3 fffbone Messages 31 Reaction score 0 Not quite the answer I was looking for, but thanks any how. I already know how the problem is solved.
Not quite the answer I was looking for, but thanks any how. I already know how the problem is solved.
Jan 10, 2004 #4 master_coda Messages 591 Reaction score 0 You asked me how to find the derivative. I showed you a way to do it. What were you looking for?
Jan 11, 2004 #5 fffbone Messages 31 Reaction score 0 I was looking for something more like this: x'=y^(1/y)*(1/y^2-ln(y)/y^2). I solved it though, guess just got stuck for a minute.
I was looking for something more like this: x'=y^(1/y)*(1/y^2-ln(y)/y^2). I solved it though, guess just got stuck for a minute.
Jan 11, 2004 #6 master_coda Messages 591 Reaction score 0 Usually people want the derivative of y wrt x. Plus, you gave an equation for y in terms of x and y.
Jan 11, 2004 #7 fffbone Messages 31 Reaction score 0 Since y does not equal f(x), y can not be expressed in terms of x. We would have to express x in terms of y. Of course, it would be much better if the equation was x=y^x, then it can be expressed as y=x^(1/x).
Since y does not equal f(x), y can not be expressed in terms of x. We would have to express x in terms of y. Of course, it would be much better if the equation was x=y^x, then it can be expressed as y=x^(1/x).