Question

if u =3(ax+by+cz)^2-(x^2+y^2+z^2) and a^2+b^2+c^2=1 show that d^2u/dx^2+d^2u/dy^2+d^2u/dy^2=0​

Answer 1

Answer:

$6 a^2-2+u\left(6 b^2-2\right)+6 b^2-2=0$

Step-by-step explanation:

A function's varied rate of change with respect to an independent variable is referred to as a derivative. When there is a variable quantity and the rate of change is irregular, the derivative is most frequently utilised. The derivative is used to assess how sensitive a dependent variable is to an independent variable (independent variable).

Differentiation is the method used to find the derivative. Anti-differentiation is the term for the reverse process. Let's determine a function's derivative, y = f. (x). It measures how quickly the value of y changes as a function of the change in the variable x. The derivative of the function "f" with respect to the variable "x" is what it is called.

The derivative of y with respect to x is expressed as dy/dx if an infinitesimal change in x is designated as dx.

Here, "dy by dx" or "dy over dx" is used to denote the derivative of y with respect to x.

$\frac{d^2}{d x^2}\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right)+$$\frac{d^2}{d y^2}\left(u\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right)\right)+$ $\frac{d^2}{d y^2}\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right.$

=$\frac{d^2}{d x^2}\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right)+\frac{d^2}{d y^2}\left(u\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right)\right)$$+\frac{d^2}{d y^2}\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right.$

$\frac{d^2}{d x^2}\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right)=6 a^2-2$

$6 a^2-2+\frac{d^2}{d y^2}\left(u\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right)\right)+\frac{d^2}{d y^2}\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right)=0$

$\frac{d^2}{d y^2}\left(u\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right)\right)=u\left(6 b^2-2\right)$

$\begin{aligned} & 6 a^2-2+u\left(6 b^2-2\right)+\frac{d^2}{d y^2}\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right)=0 \\ & \frac{d^2}{d y^2}\left(3(a x+b y+c z)^2-\left(x^2+y^2+z^2\right)\right)=6 b^2-2 \\ & 6 a^2-2+u\left(6 b^2-2\right)+6 b^2-2=0\end{aligned}$

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