IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - Differentiate-

Arjun

hiiiiiiiiiii...........

heres da soln.............

y/b=tan^-1[(x/a) + tan^-1(y/x)]

so.........

tany/b=x/a+tan^-1(y/x)

so differentiating............

sec²y/b 1/b dy/dx =1/a+1/1+(y/x)² d/dx(y/x)

(1/bsec²y/b)dy/dx =1/a + 1/1+(y/x)^2 [y+xdy/dx)

(1/bsec²y/b)dy/dx-1/1+(y/x)^2(xdy/dx)=1/a +y/1+(y/x)²

dy/dx[(1/bsec²y/b)-x/1+(y/x)²]=1/a +y/1+(y/x)²

so therefore............

dy/dx=1/a +y/1+(y/x)² /[(1/bsec²y/b)-x/1+(y/x)²]

hope u got it........

nice problem........