જો {z^2} = frac{{x^{1/2} + y^{1/2}}}{{x^{1/3} | vedclass

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(D) ધારો કે $f(x, y) = z^2 = \frac{x^{1/2} + y^{1/2}}{x^{1/3} + y^{1/3}}$.સમપરિમાણીયતા તપાસતા: $f(tx, ty) = \frac{(tx)^{1/2} + (ty)^{1/2}}{(tx)^{1/3}

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$\frac{z}{12}$

Vedclass · Answer

(D) ધારો કે $f(x, y) = z^2 = \frac{x^{1/2} + y^{1/2}}{x^{1/3} + y^{1/3}}$.સમપરિમાણીયતા તપાસતા: $f(tx, ty) = \frac{(tx)^{1/2} + (ty)^{1/2}}{(tx)^{1/3} + (ty)^{1/3}} = \frac{t^{1/2}(x^{1/2} + y^{1/2})}{t^{1/3}(x^{1/3} + y^{1/3})} = t^{1/2 - 1/3} f(x, y) = t^{1/6} f(x, y)$.અહીં $z^2$ એ $n = \frac{1}{6}$ ઘાતવાળું સમપરિમાણીય વિધેય છે,તેથી આયલરના પ્રમેય મુજબ:$x \frac{\partial (z^2)}{\partial x} + y \frac{\partial (z^2)}{\partial y} = n z^2 = \frac{1}{6} z^2$.સાંકળના નિયમનો ઉપયોગ કરતા,$\frac{\partial (z^2)}{\partial x} = 2z \frac{\partial z}{\partial x}$ અને $\frac{\partial (z^2)}{\partial y} = 2z \frac{\partial z}{\partial y}$.આ કિંમતો સમીકરણમાં મૂકતા:$x(2z \frac{\partial z}{\partial x}) + y(2z \frac{\partial z}{\partial y}) = \frac{1}{6} z^2$.બંને બાજુ $2z$ વડે ભાગતા ($z 
eq 0$ ધારતા):$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{1}{12} z$.

જો ${z^2} = \frac{{x^{1/2} + y^{1/2}}}{{x^{1/3} + y^{1/3}}}$ હોય,તો $x\frac{{\partial z}}{{\partial x}} + y\frac{{\partial z}}{{\partial y}} = $

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