Question

find all first and second derivatives of z where
z=f(2x−3y)+g(3x−4y)
@(1,1) find all first and second derivatives of z where
z=f(2x−3y)+g(3x−4y)
@(1,1) @Mathematics

Answer 1

\[∂z/∂x = f_x (2) + g_x(3)\]
\[∂z/∂y = f_y (-3) + g_y(-4)\]
is this the right idea?

Answer 2

Yes =)

Answer 3

then the second partial derivatives are all zero/???

Answer 4

Yep. On the right track

Answer 5

do i substitute the (1,1) as (x,y) or as (f,g)

Answer 6

Since in the 1st derivative, we get constants. Derviatives of constants are 0

Answer 7

Yes as we are given x,y (1,1)

Answer 8

\[{∂z \over ∂x}=f_x(2)+g_x(3)\]\[{∂z \over ∂y}=f_y(−3)+g_y(−4)\]
\[{∂^2z \over ∂ x^2} =0\]\[{∂^2z \over ∂ y^2} =0\]\[{∂^2z \over ∂ x∂y} =0\]

is the final solution? i haven't used the @(1,1) part at all

Answer 9

should i be using implicit differentiation ?

Answer 10

In this case, we cannot.

Answer 11

So far, you have it correct

Answer 12

so far?, what else do i need to do?

Answer 13

No, actually this isn't correct.

Let's look at the first component, f(2x - 3y) = f(u(x,y)), u(x,y) = 2x - 3y

Then by the chain rule,
\[ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} = f'(u(x,y)) \ . 2 = 2 f'(2x - 3y)\]

Answer 14

So re-evaluate now the first partials for z and then the second.

Answer 15

so
\[{∂f \over ∂x} = 2 f_x (2x-3y) + 3g_x (3x-4y)\]\[{∂f \over ∂y} = -3 f_y (2x-3y) -4g_y (3x-4y)\]

Answer 16

No, be careful. Not f_x but the derivative of f.

Answer 17

Remember f and g by themselves are function R --> R.

Answer 18

not functions R^2 --> R

Answer 19

ops.
\[∂z/∂x=2f_x(2x−3y)+3g_x(3x−4y)\]
\[∂z/∂y=−3f_y(2x−3y)−4g_y(3x−4y)\]
is this what you mean james?

Answer 20

I mean the partial derivative of z wrt x is
2 f'(2x - 3y) + 3 g'(3x - 4y)

not partial derivatives of f and g.

Answer 21

so \[f_x\] is different to \[f'\]because the first is a partial derivative and second is the total derivative?

Answer 22

yes. But f is also a function on a domain of one variable and therefore the partial derivative doesn't make sense. f : R --> R. Don't be fooled by the x and y in the argument of f. f is still a 1-d function.

Answer 23

For example, what is the partial derivative wrt x of z = sin(2x - 3y). So here f = sin.

Here that partial wrt x is 2 cos(2x - 3y)

Answer 24

hmmm, i think im following

Answer 25

I don't want to clutter us up with notation, but this might be useful.
z = z(x,y). Hence z is a function R^2 --> R and it is given by
\[ z = z(x,y) = f(u(x,y)) + g(v(x,y)) \]
\[ u(x,y) = 2x - 3y \ \ \ and \ \ \ v(x,y) = 3x−4y \]
f and g are functions R --> R, u and v are functions R^2 --> R

Answer 26

\[z_{xx} = 4f''+9g''\]\[z_{xy} = -6f''-12g''\]\[z_{yy} = 9f''+16g''\]

Answer 27

Right. At (x,y) = (1,1), 2x−3y = -1 and 3x - 4y = -1. Hence evaluate f'' and g'' at -1.

E.g., \( z_{xx} = 4f''(-1) + 9g''(-1) \).

Answer 28

I think i get what you are saying about f,g being 1Dimentional now

Answer 29

Or more formally still,
\[ z_{xx}(1,1) = 4f''(-1) + 9g''(-1) \]

Answer 30

Great.

Answer 31

thank you so much (again),
legendary

Answer 32

np