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How do you differentiate #tanx+tany=1#?

Answer 1

# dy/dx = - sec^2x/sec^2y #

When we differentiate #y# wrt #x# we get #dy/dx#.

However, we cannot differentiate a non implicit function of #y# wrt #x#. But if we apply the chain rule we can differentiate a function of #y# wrt #y# but we must also multiply the result by #dy/dx#.

When this is done in situ it is known as implicit differentiation.

We have:

# tanx + tany = 1 #

Differentiate wrt #x#:

# sec^2x + sec^2ydy/dx = 0 # # :. sec^2ydy/dx = - sec^2x # # :. dy/dx = - sec^2x/sec^2y #

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Answer 2

Emma Aldridge

To differentiate tan(x) + tan(y) = 1, you need to use implicit differentiation.

Differentiate each term with respect to x, keeping in mind that y is a function of x:

d/dx(tan(x)) + d/dx(tan(y)) = d/dx(1)

Apply the chain rule for differentiating tan(y) with respect to x:

sec^2(y) * dy/dx = 0

Solve for dy/dx:

dy/dx = 0

Repeat the process for d/dx(tan(x)):

sec^2(x) * dx/dx = sec^2(x)

Now, solve for dx/dx:

dx/dx = 1

Therefore, the derivatives are:

dy/dx = 0 dx/dx = 1

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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.