Why can't I use the formula for $f(x)=x^x$ to derive $(\frac{x}{x+1})^x$?

Question

Why can't I use the formula $y=x^x$, $\dot {y}=x^x(\ln x+1)$ to derive $(\frac{x}{x+1})^x$? I've tried and it doesn't get me the correct result. Have I done any mistake? $$\Bigl[\Bigl(\frac{x}{x+1}\Bigr)^x\Bigr]'=(\frac{x}{x+1})^x\Bigl[\ln\Bigl(\frac{x}{x+1}\Bigr)+1\Bigr]\Bigl[\frac{1+x-x}{(x+1)^2}\Bigr]=\frac{\bigl(\frac{x}{x+1}\bigr)^x\Bigl[\ln\bigl(\frac{x}{x+1}\bigr)+1\Bigr]}{(1+x)^2}$$ If not, please explain why I can't use the formula.

Answer 1

Hint:

Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=\bigl(u(x)\bigr)^x$ $$\frac{f'(x)}{f(x)}=\ln\bigl(u(x)\bigr) +x\,\frac{u'(x)}{u(x)}.$$

Answer 2

Hint: I have got this here :$${\frac {1}{x+1} \left( {\frac {x}{x+1}} \right) ^{x} \left( \ln \left( {\frac {x}{x+1}} \right) x+\ln \left( {\frac {x}{x+1}} \right) +1 \right) } $$ Taking the logarithm on both sides we get $$\ln(y)=x(\ln(x)-\ln(x+1))$$ and differentiating with respect to $x$ $$\frac{1}{y}y'=\ln(x)-\ln(x+1)+x\left(\frac{1}{x}-\frac{1}{x+1}\right)$$

Answer 3

Define $z:=\frac{x}{x+1}$. Your claim is that $\frac{d}{dx}(z^x)=\frac{d}{dz}(z^z)\cdot\frac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $\frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.

Answer 4

To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.

Answer 5

What you have used is that $$\left [g(x)^x\right]'=g'(x)\cdot g(x)^x\cdot \ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$\ln f(x)=x\ln x-x\ln(x+1)$$whose first-order differentiation leads to $${f'(x)\over f(x)}=1+\ln x-\ln (x+1)-{x\over x+1}$$from which we obtain$$f'(x)=\left({x\over x+1}\right)^x\cdot \left({1\over x+1}+\ln{x\over x+1}\right)$$