Finding the Directional derivative

teng125
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Find the directional derivative of f (x, y) = 5−x2−2y2 at P (-1, -1).
anybody pls help as i don't know how to find the direction from this ques so , i can't find the directional derivative


pls help
thanx
 
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Your question seems to be incomplete; please provide the complete question as it is given in the book. The directional derivative is the rate at which a function changes at some point in a particular direction.
 
neutrino said:
The directional derivative is the rate at which a function changes at some point in a particular direction.

anybody pls help as i don't know how to find the direction from this ques so

I think that's what he's trying to figure out.

If you have absolutely no clue what direction it's in, I would guess that it's in the direction of the vector from the origin to P. Not for any particular reason, but because it's a better guess than anything else :confused:
 
One more time: the problem as stated makes no sense. teng125, please state the entire problem as given in your textbook.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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