Question

[CALCULUS III—CHAIN RULE] Find ∂w/∂s and ∂w/∂t where w = y^3 −3 x^2 y, x = e^s, y = e^t

Answer 1

\[w=y^3-3x^2y\]\[x=e^s,\ y=e^t\]

Answer 2

\[ \frac{d}{ds} w = \frac{d}{ds} y^3 - 3 \frac{d}{ds} (x^2y) \]

Answer 3

\[= 3 y^2 \frac{dy}{ds} - 3 x^2\frac{dy}{ds}- 6 x y \frac{dx}{ds} \]
now find dy/ds and dx/ds

Answer 4

\[y=e^t,\ dy/ds = 0\]\[x=e^s,\ dx/ds=e^s\]

Answer 5

looks good.
\[ \frac{dw}{ds} = -6xye^s \]

Answer 6

sub in for x and y to get things in terms of s and t

Answer 7

^ This is the partial ?

If so we repeat a similar process for ∂w/∂t

Answer 8

yes, repeat for dw/dt

Answer 9

\[∂w/∂s=-6xye^s=-6e^{2s}e^t\]

Answer 10

or 2s+t for the exponent

Answer 11

^ sounds better, I did not think to combine the exponents.

Very nice.