0
$\begingroup$

Hello all math geniuses out there here is a question that probably does require much thinking, but i can't seem to figure out.

The equation $x^3 +yz-xz^2=0$ implicitly defines a function $z=f(x,y)$ for which $f(-2,1)=2$. Calculate $f_x(-2,1)$

Here is how i think about it $$f_x= 3x^2-z^2$$ $$f_y= z$$ $$f_Z=y-2xz$$ should I just evaluate everything at $(-2,1)$ or what?


Honestly every answer below is satisfying and i dont know who to give best answer for so i am just gonna give it to the first one. The rest thank you for your time

$\endgroup$
  • $\begingroup$ How is $f(-2,1)=2$? $\endgroup$ – GoodDeeds Mar 7 '16 at 12:57
2
$\begingroup$

$$ x^3+yf(x,y)-xf(x,y)^2=0 \text{\\Differentiate with respect to x}\\ 3x^2+yf_x(x,y)-f(x,y)^2-2xf(x,y)f_x(x,y)=0 $$ Plug in (-2,1): $$ 12+1\cdot{f_x(-2,1)}-4+8f_x(-2,1)=0\\ 9f_x(-2,1)=-8\\ f_x=-\frac{8}{9} $$

$\endgroup$
1
$\begingroup$

Rewrite the equation as

$x^3 + y*f(x,y) -x*f^2(x,y) = 0$

Take derivative wrt x:

$3x^2 + y*f_x(x,y) - (1*f^2(x,y) + x*2*f(x,y)*f_x(x,y)) = 0$

Now you can calculate $f_x(-2,1)$

$\endgroup$
1
$\begingroup$

The equation is quadratic in $z$, there are two possible values of $z$ when plugging in for the derivatives.

$$ 2 z = y/x \pm \sqrt{(y/x)^2+ 4 x^2} $$

and evaluate everything at:

$$ (-2,1,z_1), (-2,1,z_2) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.