How do you find tangent line equation for #1+ln(xy)=e^(x-y)# in point (1,1)?
Hence, diff. both sides of the given eqn. of curve,
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To find the tangent line equation for the given equation at the point (1,1), we need to follow these steps:
- Differentiate the equation with respect to x using the rules of differentiation.
- Substitute the values of x and y with the coordinates of the given point (1,1).
- Solve the resulting equation for the derivative to find the slope of the tangent line.
- Use the point-slope form of a line to write the equation of the tangent line.
Let's go through these steps:
-
Differentiating the equation with respect to x: d/dx [1 + ln(xy)] = d/dx [e^(x-y)]
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Substituting the values of x and y with 1: d/dx [1 + ln(1*1)] = d/dx [e^(1-1)]
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Simplifying the equation: d/dx [1 + ln(1)] = d/dx [e^0] 0 + d/dx [ln(1)] = 0
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The derivative is zero, indicating that the slope of the tangent line is zero. Therefore, the equation of the tangent line is simply the equation of the horizontal line passing through the point (1,1): y = 1
Thus, the tangent line equation for 1 + ln(xy) = e^(x-y) at the point (1,1) is y = 1.
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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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