Differentiation using logarithms.

Question

the variables $x$ and $y$ are positive and related by $$x^a\cdot y^b=(x+y)^{(a+b)}$$ where $a$ and $b$ are positive constants. By taking logarithms of both sides, show that $\frac{dy}{dx}=\frac{y}{x}$. provided that $bx$ not equal to $ay$.

Answer 1

$$a\ln x+b\ln y=(a+b)\ln(x+y)$$

Differentiating both sides wrt $x,$ $$\dfrac ax+\dfrac by\dfrac{dy}{dx}=\dfrac{a+b}{x+y}\left[1+\dfrac{dy}{dx}\right]$$

as $\dfrac{d[\ln(x+y)]}{dx}=\dfrac{d[\ln(x+y)]}{d(x+y)}\cdot\dfrac{d(x+y)}{dx}=\cdots$

$$\implies\dfrac{dy}{dx}\left[\dfrac{a+b}{x+y}-\dfrac by\right]=\dfrac ax-\dfrac{a+b}{x+y}$$

$$\iff\dfrac{dy}{dx}\cdot\dfrac{ay-bx}{y(x+y)}=\dfrac{ay-bx}{x(x+y)}$$

Answer 2

We have: $\left(\dfrac{x}{x+y}\right)^{a}\cdot \left(\dfrac{y}{x+y}\right)^{b}=1\Rightarrow a\ln\left(\dfrac{x}{x+y}\right)+b\ln\left(\dfrac{y}{x+y}\right)=0 \Rightarrow\dfrac{a(x+y)}{x}\left(\dfrac{x+y-x(1+y')}{(x+y)^2}\right)+\dfrac{b(x+y)}{y}\left(\dfrac{y'(x+y)-y(1+y')}{(x+y)^2}\right)=0\Rightarrow ay(y-xy')+bx(y'x-y)=0\Rightarrow (ay-bx)(xy'-y)=0\Rightarrow xy'-y=0\Rightarrow y'=\dfrac{y}{x}$, since $ay \neq bx$.

Answer 3

Take the logarithmic differential $$\frac axdx+\frac bydy=\frac{a+b}{x+y}(dx+dy),$$ and multiply by $xy(x+y)$, known to be nonzero: $$ay(x+y)dx+bx(x+y)dy=(a+b)xy(dx+dy),$$ or after regrouping $$x(ay-bx)dy=y(ay-bx)dx.$$