Question

please help with this...

Consider the function z(x, y) = ye^(2x^3−5) at (x, y) = (1, 3). Use the property of partial derivative to find how the number of units x must approximately change, if y increases by 0.5 unit
so that z increases by 1 unit.

i got this till now

Z'x=y.6.e^2x^3-5
z'y= y.e^(2x^3-5)

Answer 1

\[z'_{y} = e ^{2x^3-5}\] my bad

Answer 2

\[\large \mathbb{ \Delta x = x_y\Delta y+x_z \Delta z}\]

Answer 3

thats the approximation formula right ?

Answer 4

i have no idea tbh =/

Answer 5

have u seen total differential formula :
\[\large df = \dfrac{\partial f}{\partial x}dx + \dfrac{\partial f}{\partial y}dy \]

?

Answer 6

that's just the total differentiation right??

Answer 7

yes

Answer 8

oh ok, so u just rewrote the total differentation to get dx ?

Answer 9

exactly! and i have replaced differentials with Delta

Answer 10

thnx! gonna try rewrite it myself to see if i get there, btw are my derivatives correct?

Answer 11

\[\large z = ye^{2x^3-5} \]

is this the function ?

Answer 12

yup

Answer 13

when u differentiate with respect to x, all other independent variables can be treated as constant :

\[\large z_x = ye^{2x^3-5} \cdot (6x^2) = 6x^2ye^{2x^3-5} \]

Answer 14

thats the partial with respect to x right ?

Answer 15

\[z= xy^2, z _{x}= x2y \]
i dont see why the y get multiplied by e^2x^3-5

Answer 16

is it because of a property of e?

Answer 17

y can be treated as constant when u differentiate with respect to x, so u can simply pull it out of derivative

Answer 18

\[\large z = ye^{2x^3-5} \]

Answer 19

treat y as constant and differentiate with respect to x

Answer 20

\[\large \dfrac{\partial }{\partial x}(z) = \dfrac{\partial }{\partial x}\left[ye^{2x^3-5} \right]\]

Answer 21

\[\large \dfrac{\partial }{\partial x}(z) = y\dfrac{\partial }{\partial x}\left[e^{2x^3-5} \right]\]

Answer 22

\[\large \dfrac{\partial }{\partial x}(z) = y\left(e^{2x^3-5} \right) \dfrac{\partial }{\partial x}\left[2x^3-5\right] \]

Answer 23

\[\large \dfrac{\partial }{\partial x}(z) = y\left(e^{2x^3-5} \right) \left(6x^2\right)\]

Answer 24

\[\large \dfrac{\partial }{\partial x}(z) = 6x^2y\left(e^{2x^3-5} \right) \]

Answer 25

see if that makes more or less sense..