Second-order partial derivatives?
By the Chain Rule, ∂z/∂s = ∂z/∂x ∂x/∂s + ∂z/∂y ∂y/∂s ........= 2s ∂z/∂x + 3r ∂z/∂y. Next, ∂²z/∂r∂s = (∂/∂r) ∂z/∂s = (∂/∂r) [2s ∂z/∂x + 3r ∂z/∂y] = 2s (∂/∂r) ∂z/∂x + [3 * ∂z/∂y + 3r * (∂/∂r) ∂z/∂y], via product rule = 2s [∂²z/∂x² ∂x/∂r + ∂²z/∂y∂x ∂y/∂r] + 3 ∂z/∂y + 3r [∂²z/∂x∂y ∂x/∂s + ∂²z/∂y² ∂y/∂s], via Chain Rule = 2s [2r ∂²z/∂x² + 3s ∂²z/∂y∂x] + 3 ∂z/∂y + 3r [2s ∂²z/∂x∂y + 3r ∂²z/∂y²] = 4rs ∂²z/∂x² + (6s² + 6rs) ∂²z/∂y∂x + 9r² ∂²z/∂y² + 3 ∂z/∂y, assuming equality of mixed partials. I hope this helps!
Second order partial derivatives?
When you take a partial derivative with respect to a particular independent variable, you assume that all the other independent variables are held constant. So, for example, if z = x^2 + 2xy + y^3, and we want ∂z/∂x, then we can take the derivative with respect to x of each term. For the first term, no problem. For the second term, we rearrange it to (2y)x, where 2y is held constant, and the derivative with respect to x is just 2y. For the third term, with y constant, the derivative with respect to x is 0. So ∂z/∂x = 2x + 2y + 0 = 2x + 2y Similarly, if you take another partial derivative with respect to x, ∂/∂x (∂z/∂x) = ∂^2 z / ∂x^2 = ∂/∂x (2x + 2y) = 2 + 0 = 2 On the other hand, if you go back to ∂z/∂x and take the partial derivative with respect to y, you get ∂/∂y (∂z/∂x) = ∂^2 z / ∂x∂y = ∂/∂y (2x + 2y) = 0 + 2 = 2 It turns out that it doesn't matter which order you perform the differentiation in, as long as both derivatives are continuous. If we start with ∂z/∂y = 0 + 2x + 3y^2 = 2x + 3y^2 And we take the partial with respect to x, we get ∂/∂x (∂z/∂y) = ∂^2 z / ∂y∂x = ∂/∂x (2x + 3y^2) = 2 + 0 = 2 (That's not a proof, but it is provable. This is Clairaut's theorem, http://en.wikipedia.org/wiki/Symmetry_of... ) I hope that helps!
Second-Order Partial Derivatives - HELP!?
If you calculate partial derivative with respect to one variable, you treat all the other variables as constants.
A partial derivative is found by taking a normal derivative whilst holding other variables constant:[math]f(x,y) = y^{2}x + x ^{2}y[/math][math]\left.\frac{\partial f}{\partial x}\right|_y = y^2 + 2xy[/math][math]\left.\frac{\partial f}{\partial y}\right|_x = 2yx + x^2[/math]Now to find the second order partial derivative, you repeat the process. Taking the derivative with respect to one variable and holding the other constant. It is good to be explicit about which variable is being held constant, this can be done with either parentheses with a subscript or a vertical bar - [math]\left(\frac{\partial f}{\partial x}\right)_y[/math] and [math]\left.\frac{\partial f}{\partial x} \right|_y[/math]:[math]\frac{\partial}{\partial x}\frac{\partial f}{\partial x} = \frac{\partial^2 f}{\partial x^2} = 2y[/math]You can also take the partial derivative with respect to another variable to get a mixed partial derivative:[math] \frac{\partial}{\partial y} {\frac{\partial f}{\partial x}} = \frac{\partial^2 f}{\partial x \partial y} = 2y + 2x[/math]
Calculate all four second-order partial derivatives of f(x,y)= 5x^2(y) + 4xy^3?
i write it but...
Find all the first- and second-order partial derivatives of the function?
fx= 2(x+y-2)+2(x^2+y-2)*2x fxx= 2 +12x^2 fy= 2(x+y-2)+2(x^2+y-2) fyy= 2 +2 =4 fxy= 2+4x fx-fy= 2(x^2+y^2-2) (2x-1)=0 x=1/2 2(1/2+y-2)+2(1/4+y-2)=0 1+2y-4+1/2+2y-4=0 4y=15/2 so(1/2,15/8 is a critical point) fxx*fyy-fxy^2= 5*4-16>0 minimum as fxx >0 There can bee other critical points solving the system fx=0 fy=0
Find all the second-order partial derivatives of the function f(x,y) = 2x + 3y + 8xy?
First order partial derivatives. f_x = 2 + 8y f_y = 3 + 8x Second order partial derivatives: f_xx = 0 f_yy = 0 f_xy = 8 f_yx = 8 As you can see f_xy = f_yx, this is an important result in multivariable calculus.
The method is the same, irrespective of whether you are computing a total or partial derivative, the only difference is that in the latter, other arguments are kept constant.Given f(x,…), the first derivative function isg(x,…) = (f(x+dx,…)-f(x,…))/dx,Its derivative is the second derivative of f:h(x,…) = (g(x+dx,…)-g(x,…))/dx = (f(x+2dx,…)-2f(x+dx,…)+f(x,…))/dx^2Of course, in all the above, there is an implicit limit as dx -> 0.
Calculus: First and Second Partial Derivatives?
1) derivate of arctanx= 1/(x²+1), so f_x= z* 1/(y²/x² +1) * (-y/x²) = -yz/x² * x²/(y²+x²) = -yz / (y²+x²) so f_x(2,2,-4) = -2*-4 / (4+4) = 8/8 = 1 f_y=z* 1/(y²/x² +1) * (1/x) = z/x * x²/(y²+x²) = zx/(y²+x²) f_y(2,2,-4) = -4*2 / 8 = -8/8 = -1 f_z = arctan(y/x) f_z(2,2,-4) = arctan(1) = Pi/4 2) f(x,y) = -2sin(2x+y) +2cos(x-y) f_x = -2cos(2x+y)*2 + 2*-sin(x-y)*1 = -4cos(2x+y) -2sin(x-y) f_y = -2cos(2x+y)*1 + 2*-sin(x-y)*-1 = -2cos(2x+y) +2sin(x-y) f_xx= -4*-sin(2x+y)*2 -2cos(x-y)*1 = 8sin(2x+y) -2cos(x-y) f_yy= -2*-sin(2x+y)*1 +2cos(x-y)*-1 = 2sin(2x+y) -2cos(x-y) f_xy= -4*-sin(2x+y)*1 -2cos(x-y)*-1 = 4sin(2x+y) +2cos(x-y) f_yx=f_xy