Bagaimana Anda membedakan secara implisit -y ^ 2 = e ^ (2x-4y) -2yx?

Bagaimana Anda membedakan secara implisit -y ^ 2 = e ^ (2x-4y) -2yx?
Anonim

Menjawab:

# dy / dx = ((e ^ (x-2y)) ^ 2-y) / (2 (e ^ (x-2y)) ^ 2 + x-y) #

Penjelasan:

Kita dapat menulis ini sebagai:

# 2yx-y ^ 2 = (e ^ (x-2y)) ^ 2 #

Sekarang kita ambil # d / dx # dari setiap istilah:

# d / dx 2yx -d / dx y ^ 2 = d / dx (e ^ (x-2y)) ^ 2 #

# 2yd / dx x + xd / dx 2y -d / dx y ^ 2 = 2 (e ^ (x-2y)) d / dx e ^ (x-2y) # #

# 2yd / dx x + xd / dx 2y -d / dx y ^ 2 = 2 (e ^ (x-2y)) d / dx x-2y e ^ (x-2y) #

# 2yd / dx x + xd / dx 2y -d / dx y ^ 2 = 2 (e ^ (x-2y)) e ^ (x-2y) (d / dx x -d / dx 2thn) #

# 2y + xd / dx 2y -d / dx y ^ 2 = 2 (e ^ (x-2y)) ^ 2 (1-d / dx 2y) #

Menggunakan aturan rantai yang kita dapatkan:

# d / dx = dy / dx * d / dy #

# 2y + dy / dxxd / dy 2y -dy / dxd / dy y ^ 2 = 2 (e ^ (x-2y)) ^ 2 (1-dy / dxd / dy 2y) #

# 2y + dy / dx2x-dy / dx2y = 2 (e ^ (x-2y)) ^ 2 (1-dy / dx2) #

# 2y + dy / dx2x-dy / dx2y = 2 (e ^ (x-2y)) ^ 2-dy / dx4 (e ^ (x-2y)) ^ 2 #

# dy / dx4 (e ^ (x-2y)) ^ 2 + dy / dx2x-dy / dx2y = 2 (e ^ (x-2y)) ^ 2-2y #

# dy / dx (4 (e ^ (x-2y)) ^ 2 + 2x-2y) = 2 (e ^ (x-2y)) ^ 2-2y #

# dy / dx = (2 (e ^ (x-2y)) ^ 2-2y) / (4 (e ^ (x-2y)) ^ 2 + 2x-2y) = ((e ^ (x-2y))) ^ 2-y) / (2 (e ^ (x-2y)) ^ 2 + xy) #