How do you find the equation of the line tangent to #y=sinx# at #(pi/4, sqrt(2)/2)#?
The slope of the tangent line can be found using the derivative of the function:
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To find the equation of the line tangent to the curve y = sin(x) at the point (π/4, √2/2), we need to determine the slope of the tangent line at that point.
The slope of the tangent line can be found by taking the derivative of the function y = sin(x) with respect to x.
The derivative of sin(x) is cos(x).
Evaluating cos(x) at x = π/4, we get cos(π/4) = √2/2.
Therefore, the slope of the tangent line at (π/4, √2/2) is √2/2.
Using the point-slope form of a linear equation, we can write the equation of the tangent line as y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Plugging in the values, we have y - √2/2 = (√2/2)(x - π/4).
Simplifying, the equation of the tangent line is y = (√2/2)x - (√2/2)(π/4) + √2/2.
This can be further simplified to y = (√2/2)x - (√2π/8) + √2/2.
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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.
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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