How do you take the derivative of #x = tan (x+y)#?
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To find the derivative of ( x = \tan(x + y) ), you can use implicit differentiation.
[ \frac{{dy}}{{dx}} = \frac{{\frac{{d}}{{dx}}(\tan(x + y))}}{{\frac{{d}}{{dx}}x}} ]
Using the chain rule and the derivative of tangent function, the derivative of (\tan(x + y)) with respect to (x) is:
[ \sec^2(x + y) \cdot (1 + \frac{{dy}}{{dx}}) ]
The derivative of (x) with respect to (x) is simply 1.
So, putting it all together:
[ 1 = \sec^2(x + y) \cdot (1 + \frac{{dy}}{{dx}}) ]
Solve for (\frac{{dy}}{{dx}}) to find the derivative.
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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.
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.
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.
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.
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