how to solve implicit function
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Implicit differentiation. In calculus, a method called implicit differentiation makes use of the chain rule to differentiate implicitly defined functions. To differentiate an implicit function y(x), defined by an equation R(x, y) = 0, it is not generally possible to solve it explicitly for y and then differentiate.
y = tan(xy)
dy/dx = sec^2(xy)*(x dy/dx + y )
=> dy/dx = x sec^2(xy) dy/dx + y sec^2(xy)
=> dy/dx (1 - x sec^2(xy)) = y sec^2(xy)
=> dy/dx = y sec^2(xy) / (1 - x sec^2(xy))
y = tan(xy)
dy/dx = sec^2(xy)*(x dy/dx + y )
=> dy/dx = x sec^2(xy) dy/dx + y sec^2(xy)
=> dy/dx (1 - x sec^2(xy)) = y sec^2(xy)
=> dy/dx = y sec^2(xy) / (1 - x sec^2(xy))
prabalsharma:
xy=tan(xy) solve
show that dy/dx = -y/x
if xy= tan(xy) show that dy/dx = -y/x. 1 Follow0. Utkarsh Gupta, added an answer, on 14/6/15. 265 helpful votes in Math. Given xy = tan (xy)-------1. Let dy/dx= A
can u solve
2mint wait
y = tan(xy)
dy/dx = sec^2(xy)*(x dy/dx + y )
=> dy/dx = x sec^2(xy) dy/dx + y sec^2(xy)
=> dy/dx (1 - x sec^2(xy)) = y sec^2(xy)
=> dy/dx = y sec^2(xy) / (1 - x sec^2(xy))
dy/dx = sec^2(xy)*(x dy/dx + y )
=> dy/dx = x sec^2(xy) dy/dx + y sec^2(xy)
=> dy/dx (1 - x sec^2(xy)) = y sec^2(xy)
=> dy/dx = y sec^2(xy) / (1 - x sec^2(xy))
thanks
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