Hvordan differentierer du implicit -y ^ 2 = e ^ (2x-4y) -2yx?

Hvordan differentierer du implicit -y ^ 2 = e ^ (2x-4y) -2yx?
Anonim

Svar:

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

Forklaring:

Vi kan skrive dette som:

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

Nu tager vi # D / dx # af hvert udtryk:

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

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

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

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

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

Ved hjælp af kædelegemet får vi:

# 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) med) ^ 2-dy / DX4 (e ^ (x-2y) med) ^ 2 #

# Dy / DX4 (e ^ (x-2y) med) ^ 2 + dy / dx2x-dy / dx2y = 2 (e ^ (x-2y) med) ^ 2-2y #

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

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