Question

if u=log(x^3+y^3-xy^2-x^2y) than prove that (d/dx + d/dy)^2u=-4/(X+y)^2​

Answer 1

Step-by-step explanation:

Please ignore the scribbling.

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Answer 2

$(\frac{du}{dx} +\frac{du}{dy} )^{2} .u$ = $\frac{(x-y)^{2} }{(x^{3} +y^{3} -xy^{2} -x^{2} y)}$.

• The derivative of a function which involves a real variable in mathematics signifies the change in the function's value or precisely, the output value with respect to the changes in its argument that is the input value. Calculus's tool that happens to be the most important one is the derivative. The velocity of an item, for instance, the derivative of its position with respect to time and it quantifies how quickly the object's position varies as time passes.
• When derivative occurs, the slope of the tangent line to the function's graph at a given input value is known as the derivative of a function of a single variable. The function closest to that input value is the one that is best approximated linearly by the tangent line.

Here, according to the given information, we are given that,

u is given as $log(x^{3} +y^{3} -xy^{2} -x^{2} y)$.

Now, differentiating with respect to x, we get,

$\frac{d}{dx}$$log(x^{3} +y^{3} -xy^{2} -x^{2} y)$

= $\frac{1}{(x^{3} +y^{3} -xy^{2} -x^{2} y)} .(3x^{2} -y^{2} -2xy)$

Now, proceeding in the similar manner, differentiating with respect to yt, we get,

$\frac{d}{dy}$$log(x^{3} +y^{3} -xy^{2} -x^{2} y)$

=$\frac{1}{(x^{3} +y^{3} -xy^{2} -x^{2} y)} .(3y^{2} -x^{2} -2xy)$

Now, considering the left hand side of the result that we need to prove,

$(\frac{du}{dx} +\frac{du}{dy} )^{2} .u$

=$(\frac{2x^{2} +2y^{2}-4xy }{(x^{3} +y^{3} -xy^{2} -x^{2} y)}} )^{2}$

=$\frac{(x-y)^{2} }{(x^{3} +y^{3} -xy^{2} -x^{2} y)}$.

Therefore, $(\frac{du}{dx} +\frac{du}{dy} )^{2} .u$ = $\frac{(x-y)^{2} }{(x^{3} +y^{3} -xy^{2} -x^{2} y)}$.

Hence, proved.

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