If tan(xy) = x, then dy/dx =
4 answers
wats your question?
d/dx (tan u) = sec^2 u du/dx
so
d/dx (tan xy) = sec^2 (xy) [x dy/dx + y]
so
sec^2(xy) [x dy/dx + y] = dx/dx = 1
sec^2(xy) x dy/dx = [1 - y sec^2(xy) ]
dy/dx = [1 - y sec^2(xy) ]/[x sec^2(xy)]
so
d/dx (tan xy) = sec^2 (xy) [x dy/dx + y]
so
sec^2(xy) [x dy/dx + y] = dx/dx = 1
sec^2(xy) x dy/dx = [1 - y sec^2(xy) ]
dy/dx = [1 - y sec^2(xy) ]/[x sec^2(xy)]
but sec^2 (xy) = 1/cos^2 (xy
so
dy/dx = [ cos^2(xy) - y ] / x
so
dy/dx = [ cos^2(xy) - y ] / x
What are all values of k for which integral x^2 dx=0