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How do you implicitly differentiate #-1=sin(x+y) #?

Answer 1

Emily Ammerman

#dy/dx= -1#

We apply the identity that if #sinx = y#, then #arcsiny = x#.

#arcsin(-1) = x + y#

The value of #arcsin(-1)#, or #270˚#, is just a constant, so the derivative will be #0#.

#d/dx(arcsin(-1)) = d/dx(x + y)#

#0 = 1 + dy/dx#

#dy/dx= -1#

Hopefully this helps!

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Answer 2

Luke Alvarez

To implicitly differentiate ( -1 = \sin(x+y) ), we'll differentiate both sides of the equation with respect to ( x ) treating ( y ) as a function of ( x ).

Apply the chain rule for the left side and the chain rule and the derivative of sine function for the right side.

The derivative of (-1) with respect to (x) is (0).

The derivative of (\sin(x+y)) with respect to (x) involves the chain rule: [ \frac{d}{dx}(\sin(x+y)) = \cos(x+y) \cdot \frac{d}{dx}(x+y) ]

[ = \cos(x+y) \cdot (1 + \frac{dy}{dx}) ]

Thus, the implicit derivative of ( -1 = \sin(x+y) ) with respect to ( x ) is:

[ 0 = \cos(x+y) \cdot (1 + \frac{dy}{dx}) ]

Solve for ( \frac{dy}{dx} ) if necessary.

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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.