# How do you find the equation of the line tangent to #y=4secx–8cosx# at the point (pi/3,4)?

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To find the equation of the line tangent to the curve y = 4sec(x) - 8cos(x) at the point (π/3, 4), we need to find the derivative of the function and evaluate it at the given point.

The derivative of y = 4sec(x) - 8cos(x) can be found using the chain rule and the derivative rules for sec(x) and cos(x).

dy/dx = 4sec(x)tan(x) + 8sin(x)

Now, substitute π/3 into the derivative expression to find the slope of the tangent line at that point.

dy/dx = 4sec(π/3)tan(π/3) + 8sin(π/3)

Simplifying, we get:

dy/dx = 4(2/√3)(√3/3) + 8(√3/2)

dy/dx = 8/√3 + 4√3

dy/dx = (8 + 4√3√3)/√3

dy/dx = (8 + 12√3)/√3

Now, we have the slope of the tangent line at the point (π/3, 4).

Using the point-slope form of a line, we can write the equation of the tangent line:

y - 4 = [(8 + 12√3)/√3](x - π/3)

Simplifying further, we get:

y - 4 = (8 + 12√3)(x - π/3)/√3

This is the equation of the line tangent to y = 4sec(x) - 8cos(x) at the point (π/3, 4).

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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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