Suppose z=x2sin(y),x=−2s2−t2,y=−6st A. Use the chain rule to find ∂s∂z and ∂t∂z as functions of x,y,s and t. ∂s∂z=−4sxsin(y)−6tx2cos(y)∂t∂z=−2txsin(y)−6sx2cos(y) B. Find the numerical values of ∂s∂z and ∂t∂z when (s,t)=(4,−5) ∂s∂z(4,−5)=−16xsin(y)+30x2cos(y) ∂t∂z(4,−5)=8xsin(y)+30x2cos(y)

Given data z=x^2sin(y), x=-2s^2-t^2,y=-6st Partial derivatives

Resuelto Suppose z=x2sin(y),x=−2s2−t2,y=−6st A. Use the chain

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Pregunta: Suppose z=x2sin(y),x=−2s2−t2,y=−6st A. Use the chain rule to find ∂s∂z and ∂t∂z as functions of x,y,s and t. ∂s∂z=−4sxsin(y)−6tx2cos(y)∂t∂z=−2txsin(y)−6sx2cos(y) B. Find the numerical values of ∂s∂z and ∂t∂z when (s,t)=(4,−5) ∂s∂z(4,−5)=−16xsin(y)+30x2cos(y) ∂t∂z(4,−5)=8xsin(y)+30x2cos(y)

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Solución
Paso 1
Given data $z = x^{2} \sin (y), x = - 2 s^{2} - t^{2}, y = - 6 s t$
Partial derivatives
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Texto de la transcripción de la imagen:

Suppose

z = x^{2} sin (y), x = - 2 s^{2} - t^{2}, y = - 6 s t

A. Use the chain rule to find

\frac{\partial z}{\partial s}

and

\frac{\partial z}{\partial t}

as functions of

x, y, s

and

t

\frac{\partial z}{\partial s} = - 4 s x sin (y) - 6 t x^{2} cos (y) \frac{\partial z}{\partial t} = - 2 t x sin (y) - 6 s x^{2} cos (y)

B. Find the numerical values of

\frac{\partial z}{\partial s}

and

\frac{\partial z}{\partial t}

when

(s, t) = (4, - 5)

\frac{\partial z}{\partial s} (4, - 5) = - 16 x sin (y) + 30 x^{2} cos (y)

\frac{\partial z}{\partial t} (4, - 5) = 8 x sin (y) + 30 x^{2} cos (y)