Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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)]

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.