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MIT 18.01 Single Variable Calculus (OCW) 59 Online
OpenStudy (anonymous):

Could someone please explain, in the implicit differentiation example [ x^2 + y^2 = 1 ], why do I have to apply the chain rule to y?

OpenStudy (anonymous):

Could you write the exercise?

OpenStudy (anonymous):

The chain rule must be used whenever the function y is being differentiated because of our assumption that y may be expressed as a function of x.

OpenStudy (anonymous):

Yes, (y) is a function of (x). Imagine that y= x^2+1 then: y^2= (x^2+1)^2, now if you apply the chain rule to both sides you get: 2y(Dy)= (2)(2x) (x^2+1). That is it. I hope you get something.

OpenStudy (anonymous):

Thanks to all you, guys. Now it's much more easier to understand the chain rule. Thank you

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