1 = e ^ y-xcos (xy)의 함축적 인 미분은 무엇입니까?

대답:

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

설명:

# 1 = e ^ y-xcos (xy) #

#rArr (d1) / dx = d / dx (e ^ y-xcos (xy)) #

# rArr0 = (de ^ y) / dx- (d (xcos (xy))) / dx #

# rArr0 = (dy / dx) e ^ y - (((dx) / dx) cosxy + x (dcosxy) / dx) #

# rArr0 = (dy / dx) e ^ y- (cosxy + x (dxy) / dx (-sinxy)) #

(cosx + x (y + x (dy) / dx) (- sinxy))) # rArr0 = (dy / dx)

(ryr0) = (dy / dx) ey- (cosxy + x (-ysinxy-x (dy) / dx (sinxy)))

# rArr0 = (dy / dx) e ^ y- (cosxy-xysinxy-x ^ 2 (dy) / dx (sinxy)) #

# rArr0 = (dy / dx) e ^ y-cosxy + xysinxy + x ^ 2 (dy) / dx (sinxy) #

# rArr0 = (dy / dx) e ^ y + x ^ 2 (dy) / dx (sinxy) -cosxy + xysinxy #

# rArr0 = (dy / dx) (e ^ y + x ^ 2 (sinxy)) - cosxy + xysinxy #

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

#rArr (dy) / dx = (cosxy-xysinxy) / (e ^ y + x ^ 2 (sinxy)) #