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How do you implicitly differentiate #9=sin^2x-cos^2y#?

Answer 1

#dy/dx=-(sinxcosx)/(sinycosy)#

Recall that when implicitly differentiating, any derivatives of terms with #y# will spit out a #dy/dx# term thanks to the chain rule.

#d/dx[9=sin^2x-cos^2y]#

#0=2sinx*d/dx[sinx]-2cosy*d/dx[cosy]#

#0=2sinxcosx-2cosy(-siny)dy/dx#

#dy/dx=-(sinxcosx)/(sinycosy)#

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Answer 2

Paisley Johnson

To implicitly differentiate (9 = \sin^2(x) - \cos^2(y)), differentiate each term with respect to (x) and (y) respectively, using the chain rule when necessary. Then solve for (\frac{{dy}}{{dx}}).

The steps are:

Differentiate (\sin^2(x)) with respect to (x).
Differentiate (\cos^2(y)) with respect to (y).
Use implicit differentiation rules to differentiate (9) with respect to both (x) and (y).
Solve for (\frac{{dy}}{{dx}}).

Here are the steps:

(\frac{{d}}{{dx}}(\sin^2(x)) = 2\sin(x)\cos(x))
(\frac{{d}}{{dy}}(\cos^2(y)) = -2\cos(y)\sin(y))
(\frac{{d}}{{dx}}(9) = 0) and (\frac{{d}}{{dy}}(9) = 0)
Combine the derivatives and solve for (\frac{{dy}}{{dx}}).

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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.