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Consider the functions z = 4 e^x ln y, x = ln (u cos v), and y = u sin v.

Express dz/du and dz/dv as functions of u and y both by using the Chain Rule and by expressing z directly in terms of u and v before differentiating.

Sagot :

facundo3141592

Answer:

remember the chain rule:

h(x) = f(g(x))

h'(x) = f'(g(x))*g'(x)

or:

dh/dx = (df/dg)*(dg/dx)

we know that:

z = 4*e^x*ln(y)

where:

y = u*sin(v)

x = ln(u*cos(v))

We want to find:

dz/du

because y and x are functions of u, we can write this as:

dz/du = (dz/dx)*(dx/du) + (dz/dy)*(dy/du)

where:

(dz/dx)  = 4*e^x*ln(y)

(dz/dy) = 4*e^x*(1/y)

(dx/du) = 1/(u*cos(v))*cos(v) = 1/u

(dy/du) = sin(v)

Replacing all of these we get:

dz/du = (4*e^x*ln(y))*( 1/u) + 4*e^x*(1/y)*sin(v)

          = 4*e^x*( ln(y)/u + sin(v)/y)

replacing x and y we get:

dz/du = 4*e^(ln (u cos v))*( ln(u sin v)/u + sin(v)/(u*sin(v))

dz/du = 4*(u*cos(v))*(ln(u*sin(v))/u + 1/u)

Now let's do the same for dz/dv

dz/dv = (dz/dx)*(dx/dv) + (dz/dy)*(dy/dv)

where:

(dz/dx)  = 4*e^x*ln(y)

(dz/dy) = 4*e^x*(1/y)

(dx/dv) = 1/(cos(v))*-sin(v) = -tan(v)

(dy/dv) = u*cos(v)

then:

dz/dv = 4*e^x*[ -ln(y)*tan(v) + u*cos(v)/y]

replacing the values of x and y we get:

dz/dv = 4*e^(ln(u*cos(v)))*[ -ln(u*sin(v))*tan(v) + u*cos(v)/(u*sin(v))]

dz/dv = 4*(u*cos(v))*[ -ln(u*sin(v))*tan(v) + 1/tan(v)]

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