How do you implicitly differentiate #-1=xysin(x/y) #?

1 Answer
Mar 19, 2016

#xcos(x/y)-x^2/ycos(x/y)(dy)/(dx)+x(dy)/(dx)sin(x/y)+ysin(x/y)=0#

Explanation:

Implicit differentiation is used when a function, here #y#, is not explicitly written in terms #x#. One will need the formula for differentiation of product & ratios of functions as well as chain formula to solve the given function.

For this, we take differential of both sides and using product formula gives us

#0=xyd/(dx)sin(x/y)+x(dy)/(dx)sin(x/y)+ysin(x/y)#

Now differentiating #d/(dx)sin(x/y)# in first term gives

#cos(x/y)d/(dx)(x/y)=cos(x/y)[x*d/(dx)(1/y)+1/y]#

= #cos(x/y)[x*(-1/y^2)(dy)/(dx)+1/y]#

= #cos(x/y){1/y-x/y^2(dy)/(dx)}#

Hence derivative of given function is

#xy[cos(x/y){1/y-x/y^2(dy)/(dx)}]+x(dy)/(dx)sin(x/y)+ysin(x/y)=0# or

#xcos(x/y)-x^2/ycos(x/y)(dy)/(dx)+x(dy)/(dx)sin(x/y)+ysin(x/y)=0#