Q1)
x³ + y³ = 3
The function is in implicit form.
Differentiating with respect to x
3x² + 3y² dxdy=0
=> 3y² dxdy = -3x²
=> y² dxdy= -x²
=> dxdy=−y2x2
Q2
y = (sin x)tan x
Taking logarithm of both sides
logey = loge(sin x)tan x
=> logey =tan x loge(sin x)
Differentiating with respect to x
y1dxdy=tanxdxdlogesinx+logesinxdxdtanx
y1dxdy=tanxsinx1cosx+logesinx.(sec2x)
=> y1dxdy=tanx.cotx+logesinx.(sec2x)
=> dxdy = y (1 +sec²x logesinx)
=> dxdy = (sin x)tan x(1 + sec²x logesinx)
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