Find all the first and second order partial derivatives of the function f(x,y) = 2x^3+ 3x^2 y^2 + 3x + 2/y + 4
Well well, looks like we have a fancy function over here. Let's dive in and find those partial derivatives, shall we?
First, let's find the first order partial derivatives. We'll start with the derivative with respect to x:
f_x = d/dx (2x^3+ 3x^2 y^2 + 3x + 2/y + 4)
Now, when we differentiate x^n, we bring down the exponent as a coefficient and lower the exponent by 1. So, applying that rule, we get:
f_x = 6x^2 + 6xy^2 + 3
Alright, now let's have some fun and find the derivative with respect to y. Here we go:
f_y = d/dy (2x^3+ 3x^2 y^2 + 3x + 2/y + 4)
Now, the fun part is differentiating with respect to y. When we differentiate y^n, we bring down the exponent as a coefficient and lower the exponent by 1, just like before. So, applying that rule, we get:
f_y = 6x^2y + 4/y^2
Voila! We found the first order partial derivatives. Now let's move on to the second order.
For the second order partial derivatives, we need to differentiate the first order derivatives we just found.
First, the second order derivative with respect to x:
f_xx = d/dx (f_x) = d/dx (6x^2 + 6xy^2 + 3)
When we differentiate x^n, we bring down the exponent as a coefficient, lower the exponent by 1. Let's do that:
f_xx = 12x + 6y^2
Alright, now let's find the second order derivative with respect to y:
f_yy = d/dy (f_y) = d/dy (6x^2y + 4/y^2)
Now, when we differentiate y^n, we bring down the exponent as a coefficient and lower the exponent by 1. Let's do that:
f_yy = 6x^2 - 8/y^3
Oh, this math stuff can get complex, just like life! But with some patience and humor, we can crack these partial derivatives together. Keep up the good work, and let me know if you have any more questions!
To find the first and second order partial derivatives of the function f(x, y) = 2x^3 + 3x^2y^2 + 3x + 2/y + 4, we take the partial derivatives with respect to each variable.
First-order partial derivatives:
∂f/∂x = ∂(2x^3 + 3x^2y^2 + 3x + 2/y + 4)/∂x
∂f/∂x = 2(3x^2) + 2(3y^2)(∂x/∂x) + 3(∂x/∂x) + 0 + 0
= 6x^2 + 6y^2 + 3
∂f/∂y = ∂(2x^3 + 3x^2y^2 + 3x + 2/y + 4)/∂y
∂f/∂y = 2(3x^2)(∂y/∂y) + 2y^2(∂y/∂y) + 0 + 2(1/y^2)(-1/y^2) + 0
= 6x^2 + 2y^2/y^2
= 6x^2 + 2
Second-order partial derivatives:
∂^2f/∂x^2 = ∂(6x^2 + 6y^2 + 3)/∂x
∂^2f/∂x^2 = 12x + 0 + 0
= 12x
∂^2f/∂y^2 = ∂(6x^2 + 2)/∂y
∂^2f/∂y^2 = 0 + 0
= 0
∂^2f/∂x∂y = ∂(6x^2 + 2)/∂y
∂^2f/∂x∂y = 0 + 0
= 0
Therefore, the first-order partial derivatives are:
∂f/∂x = 6x^2 + 6y^2 + 3
∂f/∂y = 6x^2 + 2
And the second-order partial derivatives are:
∂^2f/∂x^2 = 12x
∂^2f/∂y^2 = 0
∂^2f/∂x∂y = 0
To find the first and second-order partial derivatives of the function f(x, y) = 2x^3 + 3x^2y^2 + 3x + 2/y + 4, we need to differentiate the function with respect to each variable.
First-order partial derivatives:
To find the first-order partial derivative with respect to x, we differentiate the function with respect to x while treating y as a constant.
∂f/∂x = ∂/∂x (2x^3 + 3x^2y^2 + 3x + 2/y + 4)
Differentiating each term with respect to x, we get:
∂f/∂x = (6x^2) + (6xy^2) + 3 + 0 - (2/y^2)
Simplifying the expression, we have:
∂f/∂x = 6x^2 + 6xy^2 + 3 - (2/y^2)
Similarly, to find the first-order partial derivative with respect to y, we differentiate the function with respect to y while treating x as a constant.
∂f/∂y = ∂/∂y (2x^3 + 3x^2y^2 + 3x + 2/y + 4)
Differentiating each term with respect to y, we get:
∂f/∂y = 0 + (6x^2y) + 0 - (2/y^2) + 0
Simplifying further, we have:
∂f/∂y = 6x^2y - (2/y^2)
Second-order partial derivatives:
To find the second-order partial derivatives, we differentiate the first-order partial derivatives with respect to the corresponding variable.
∂^2f/∂x^2 = ∂/∂x (6x^2 + 6xy^2 + 3 - (2/y^2))
Differentiating ∂f/∂x (which we already found) with respect to x, we get:
∂^2f/∂x^2 = 12x + 6y^2
Similarly, we can find ∂^2f/∂y^2 by differentiating ∂f/∂y (which we already found) with respect to y:
∂^2f/∂y^2 = 6x^2 + 4/y^3
To find the mixed second-order derivative ∂^2f/∂x∂y, we differentiate ∂f/∂y (which we already found) with respect to x:
∂^2f/∂x∂y = 12xy
Finally, the mixed second-order derivative ∂^2f/∂y∂x can be found by differentiating ∂f/∂x (which we already found) with respect to y:
∂^2f/∂y∂x = 12xy
Note that in this case, the mixed partial derivatives are equal since they involve different orders of differentiation.
just plug and chug. Is there some reason you didn't show your work?
Fx = 6x + 6xy^2 + 3
Fy = 6x^2 y - 2/y^2
Fxx = 6 + 6y^2
Fxy = 12xy
Fyy = 6x^2 + 4/y^3