How do you find an equation of the tangent line to the curve at the given point: #y=(2x)/(x+1)^2# at point (0,0)?
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To find the equation of the tangent line to the curve at the point (0,0), we need to find the derivative of the function y=(2x)/(x+1)^2 and evaluate it at x=0. The derivative of the function is given by the quotient rule, which states that the derivative of (f(x)/g(x)) is (f'(x)g(x) - f(x)g'(x))/(g(x))^2. Applying the quotient rule to the given function, we get:
y' = [(2(x+1)^2 - 2x(2(x+1))]/((x+1)^2)^2
Simplifying this expression, we have:
y' = [2(x+1)^2 - 4x(x+1)]/[(x+1)^2]^2
Evaluating this expression at x=0, we get:
y'(0) = [2(0+1)^2 - 4(0)(0+1)]/[(0+1)^2]^2
Simplifying further, we have:
y'(0) = [2(1) - 4(0)]/[1]^2
y'(0) = 2/1
y'(0) = 2
Therefore, the slope of the tangent line at the point (0,0) is 2.
Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values x1=0, y1=0, and m=2 into the equation to find the equation of the tangent line:
y - 0 = 2(x - 0)
Simplifying this equation, we have:
y = 2x
Therefore, the equation of the tangent line to the curve y=(2x)/(x+1)^2 at the point (0,0) is y = 2x.
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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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