Question

Z=2x^3 * y+e^2x+3y +lny. dz/dx, dz/dy-?​

Answer 1

Ответ:

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Пошаговое объяснение:

$z=2x^3y+e^{2x}+3y+lny\\\\\frac{dz}{dx}=2y*3x^2+e^{2x}*2 =6x^2y+2e^{2x}\\\frac{dz}{dy} =2x^3+3+\frac{1}{y}$

Answer 2

To find the partial derivatives of the given function Z with respect to x and y, we'll differentiate each term separately and then combine the results.

Given function: Z = 2x^3 * y + e^(2x+3y) + ln(y)

Let's find the partial derivative dz/dx:

1. Differentiate the first term: 2x^3 * y d/dx [2x^3 * y] = 6x^2 * y

2. Differentiate the second term: e^(2x+3y) d/dx [e^(2x+3y)] = e^(2x+3y) * (2)

3. Differentiate the third term: ln(y) d/dx [ln(y)] = (1/y)

Combine the derivatives: dz/dx = 6x^2 * y + 2 * e^(2x+3y) + (1/y)

Now, let's find the partial derivative dz/dy:

1. Differentiate the first term: 2x^3 * y d/dy [2x^3 * y] = 2x^3

2. Differentiate the second term: e^(2x+3y) d/dy [e^(2x+3y)] = e^(2x+3y) * (3)

3. Differentiate the third term: ln(y) d/dy [ln(y)] = (1/y)

Combine the derivatives: dz/dy = 2x^3 + 3 * e^(2x+3y) + (1/y)

So, the partial derivatives are: dz/dx = 6x^2 * y + 2 * e^(2x+3y) + (1/y) dz/dy = 2x^3 + 3 * e^(2x+3y) + (1/y)