# Partial Derivative Question

I am given with the function: $f(x,y) = \frac{y\ln(1+x^2 + ay^2) } {x^2 + 2y^2}$ when $(x,y)\neq (0,0)$, and $f(0,0)=0$ .

There is another given data; $f_y (0,0) = 2$ . What is the value of $a$ ?

I've tried computing the limit $\frac{f(0,h)- f(0,0)}{h}$ , but it seems like it's always zero, contradicting the fact that $f_y(0,0)=2$ !

Can someone help me understand my mistake?

Thanks !

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A number followed by an exclamation mark reminds me of factorial5! –  FrenzY DT. Jul 21 '12 at 8:42
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## 2 Answers

\begin{align} f_y(0, 0) &= \lim_{h \to 0} \frac {f(0, h) - f(0, 0)} h\\ &= \lim_{h \to 0} \frac {f(0, h)} h\\ &=\lim_{h \to 0} \frac a 2 \frac {\ln(1 + ah^2)} {ah^2}\\ &= \lim_{h \to 0} \frac a 2 \frac {\ln(1 + h)} {h}\\ &= \frac a 2 \left. \frac {d} {dx} \right |_{x=1} \ln x\\ &= \frac a 2 \end{align} Since $f_y(0, 0) = 2$, $a$ must be $4$.

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Dear @Albert: There's a fallacy in what you wrote. When substituting y=h, you don't get $\frac{a}{2}$ outside the fraction, but $\frac{h}{2 }$ ! In addition, you'll get in the denominator $2h^2$ and not $ah^2$ . –  joshua Jul 21 '12 at 8:49
@joshua: $$\displaystyle \frac {f(0, h)} h = \frac 1 h \frac {h \ln(1 + ah^2)} {2h^2} = \frac {\ln(1 + ah^2)} {2h^2} = \frac a 2 \frac {\ln(1 + ah^2)} {ah^2}$$ –  AlbertH Jul 21 '12 at 8:58
OH right... Thanks a lot !!! –  joshua Jul 21 '12 at 9:13
You are welcome! –  AlbertH Jul 21 '12 at 15:07
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What about a direct L'Hospital (=L'H)?:

$$\lim_{h\to 0}\frac{1}{h}f(0,h)=\lim_{h\to 0}\frac{1}{\rlap{/}h}\frac{\rlap{/}h\log(1+ah^2)}{2ah^2}\stackrel{\text{L'H}}=\lim_{h\to 0}\frac{\rlap{/}2a\rlap{/}h}{1+ah^2}\frac{1}{\rlap{/_2}4\,\,\,\rlap{/}h}=\frac{a}{2}$$

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