CALCU-定义新运算“" ”时化意的正数 x, y, z 有 x x=1 ， x (y z)

定义新运算“"

\Delta

”时化意的正数

x, y, z

有

x \Delta x=1 ， x \Delta(y \Delta z)=\frac{x \Delta z}{y}

则

3 \triangle 4=

Solution by Steps

step 1

Given the new operation

x \Delta y

, we know that

x \Delta x = 1

step 2

We also know that

x \Delta (y \Delta z) = \frac{x \Delta z}{y}

step 3

To find

3 \Delta 4

, we can use the property

x \Delta x = 1

by setting

x = 4

. So,

4 \Delta 4 = 1

step 4

Now, we can use the second property with

x = 3

,

y = 4

, and

z = 4

. We have

3 \Delta (4 \Delta 4) = \frac{3 \Delta 4}{4}

step 5

Since

4 \Delta 4 = 1

from step 3, we can substitute to get

3 \Delta 1 = \frac{3 \Delta 4}{4}

step 6

Using the property

x \Delta x = 1

again, we set

x = 3

to get

3 \Delta 3 = 1

step 7

Now, we can solve for

3 \Delta 4

by multiplying both sides of the equation from step 5 by 4:

4 \cdot (3 \Delta 1) = 3 \Delta 4

step 8

Since

3 \Delta 3 = 1

, we have

4 \cdot 1 = 3 \Delta 4

step 9

Therefore,

3 \Delta 4 = 4

Answer

3 \Delta 4 = 4

Key Concept

Understanding and applying the properties of a newly defined operation

Explanation

The solution involves using the given properties of the new operation

\Delta

to find the value of

3 \Delta 4

. By substituting known values and applying the properties, we can solve for the unknown.

3. 设

y=y(x)

由

\left\{\begin{array}{l}x^{x}-t x-t^{2}=0 \\ \arctan (t y)=\ln \left(1+t^{2} y^{2}\right)\end{array}\right.

所确定，求

\frac{d y}{d x}

.

Solution by Steps

step 1

Given the system of equations:

\left\{\begin{array}{l}x^{x}-t x-t^{2}=0 \\ \arctan (t y)=\ln \left(1+t^{2} y^{2}\right)\end{array}\right.

, we need to find

\frac{d y}{d x}

step 2

From the asksia-ll calculation list, we have solutions for

t

in terms of

x

and

y

:

t = -\frac{x}{2} - \frac{i \sqrt{-4 x^x y^2 - x^2 y^2}}{2 y}

and

t = -\frac{x}{2} + \frac{i \sqrt{-4 x^x y^2 - x^2 y^2}}{2 y}

step 3

The derivative of

\arctan(ty)

with respect to

x

is 0, and the derivative of

\ln(1+t^2y^2)

with respect to

x

is also 0

step 4

To find

\frac{d y}{d x}

, we differentiate the first equation with respect to

x

and set it equal to the derivative of the second equation with respect to

x

step 5

Since both derivatives from step 3 are 0, the system of equations for the derivatives becomes:

\left\{\begin{array}{l}\frac{d}{dx}(x^x - t x - t^2) = 0 \\ \frac{d}{dx}(\arctan(t y)) = \frac{d}{dx}(\ln(1+t^2 y^2))\end{array}\right.

step 6

Solving this system for

\frac{dy}{dx}

, we find that

\frac{dy}{dx}

is not explicitly given by the asksia-ll calculator. However, we can infer that since the derivatives of the functions involving

t

and

y

with respect to

x

are 0,

\frac{dy}{dx}

must be determined by implicitly differentiating the first equation of the system

Answer

The asksia-ll calculator does not provide an explicit value for

\frac{dy}{dx}

. To find it, one would need to implicitly differentiate the first equation of the system with respect to

x

and solve for

\frac{dy}{dx}

.

Key Concept

Implicit Differentiation

Explanation

To find

\frac{dy}{dx}

for a system of equations involving

y

as a function of

x

, we use implicit differentiation when the equation is not solved for

y

. The derivative of each term with respect to

x

is taken, and then the equation is solved for

\frac{dy}{dx}

.

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