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How do you find dy/dx for the function #y=sin(2x+4y)#?

Answer 1

You have an Implicit Function in which #y# is function of #x# so you have to derive it as well. You get: #dy/dx=[cos(2x+4y)]*[2+4dy/dx]# where you use the Chain Rule on the #sin# function.

#dy/dx=2cos(2x+4y)+4cos(2x+4y)dy/dx# Collecting #dy/dx#: #dy/dx[1-4cos(2x+4y)]=2cos(2x+4y)# #dy/dx=(2cos(2x+4y))/(1-4cos(2x+4y))#

Hope it helps

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Answer 2

Ezekiel Alexander

To find dy/dx for the function y = sin(2x + 4y), you would use implicit differentiation. The derivative dy/dx can be found using the chain rule and implicit differentiation techniques. The steps are as follows:

Differentiate both sides of the equation with respect to x.
Apply the chain rule to differentiate sin(2x + 4y) with respect to (2x + 4y).
Solve for dy/dx.

Following these steps, you would find that dy/dx equals (2cos(2x + 4y))/(1 - 8cos^2(2x + 4y)).

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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.