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How do you differentiate #cos y +3x^2=6y#?

Answer 1

Ezra Adams

I got: #(dy)/(dx)=(6x)/(6+sin(y))#

Here we can use implicit differentiation including the term #(dy)/(dx)# when differentiating #y#.

We get:

#-sin(y)(dy)/(dx)+6x=6(dy)/(dx)#

we can collect #(dy)/(dx)# on one side to get:

#(dy)/(dx)=(6x)/(6+sin(y))#

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Answer 2

Aurora Alvarado

To differentiate the equation ( \cos(y) + 3x^2 = 6y ) with respect to ( x ), apply implicit differentiation:

Differentiate both sides of the equation with respect to ( x ).
Use the chain rule when differentiating terms involving ( y ).
Solve for ( \frac{{dy}}{{dx}} ) after differentiation.

Differentiating ( \cos(y) + 3x^2 = 6y ) with respect to ( x ) yields:

[ -\sin(y) \frac{{dy}}{{dx}} + 6x = 6 \frac{{dy}}{{dx}} ]

Now, isolate ( \frac{{dy}}{{dx}} ) terms on one side:

[ 6 \frac{{dy}}{{dx}} - \sin(y) \frac{{dy}}{{dx}} = 6x ]

[ (6 - \sin(y)) \frac{{dy}}{{dx}} = 6x ]

Finally, solve for ( \frac{{dy}}{{dx}} ):

[ \frac{{dy}}{{dx}} = \frac{{6x}}{{6 - \sin(y)}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.