# How do you find the equation of the tangent line to the graph #y=2^-x# through point (-1,2)?

Use logarithmic differentiation.

Evaluate the derivative at

Use the point-slope form of the equation of the line.

Use the natural logarithm on both sides of the equation:

Differentiate both sides:

Multiply both sides by y:

Substitute for y:

Using the point-slope form of the equation of the line:

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To find the equation of the tangent line to the graph (y=2^{-x}) through the point (-1,2), we need to find the derivative of the function at the given point. The derivative of (y=2^{-x}) is (dy/dx = -\ln(2) * 2^{-x}). Plugging in the x-coordinate of the given point (-1) into the derivative equation gives us the slope of the tangent line at that point. So, (dy/dx = -\ln(2) * 2^{-(-1)} = -\ln(2) * 2). Thus, the slope of the tangent line at the point (-1,2) is (m = -\ln(2) * 2). Using the point-slope form of the equation of a line, we have (y - 2 = -\ln(2) * 2 * (x + 1)). Simplifying this equation gives the equation of the tangent line: (y = -\ln(2) * 2 * x + (2 - 2 * \ln(2))).

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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.

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