How do you find dy/dx in terms of x and y if (x^3)y-x-2y-6=0?
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#(3x^(2)y+x^(3)y')-1-2y'=0#
For this, you use implicit differentiation (basically regular differentiation, but with y as well as x). It works the same way as regular differentiation, except every time you have to take the derivative of y, you have to put a #y'# after it.For the first term, you have to use the product rule, which is:
#(f'(x)×g(x))+(f(x)×g'(x))#
Here #f(x)=x^3# and #g(x)=y#. So you take the derivative of #x^3# and multiply by #y#, then then switch, so you get the first term (surrounded by parentheses in the answer).
The rest is pretty simple, the derivative of #x# is #1#, the derivative of #-2y# is #-2y'# and #6# and #0# are both #0#. You put all together, and you get:
#(3x^(2)y+x^(3)y')-1-2y'=0#
Which is the answer : )
#(3x^(2)y+x^(3)y')-1-2y'=0#
For this, you use implicit differentiation (basically regular differentiation, but with y as well as x). It works the same way as regular differentiation, except every time you have to take the derivative of y, you have to put a #y'# after it.For the first term, you have to use the product rule, which is:
#(f'(x)×g(x))+(f(x)×g'(x))#
Here #f(x)=x^3# and #g(x)=y#. So you take the derivative of #x^3# and multiply by #y#, then then switch, so you get the first term (surrounded by parentheses in the answer).
The rest is pretty simple, the derivative of #x# is #1#, the derivative of #-2y# is #-2y'# and #6# and #0# are both #0#. You put all together, and you get:
#(3x^(2)y+x^(3)y')-1-2y'=0#
Which is the answer : )
