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1) Find an equation for the tangent to the curve [tex]y=\frac{8x}{x^2+1}[/tex] at the point [tex]\((1, 4)\)[/tex].


Sagot :

Sure, let's go through the process of finding the equation for the tangent line step by step.

1. Identify the given curve and the point of tangency:

The curve is given by [tex]\( y = \frac{8x}{x^2 + 1} \)[/tex], and the point of tangency is [tex]\((1, 4)\)[/tex].

2. Calculate the derivative of the curve:

The derivative of [tex]\( y = \frac{8x}{x^2 + 1} \)[/tex] with respect to [tex]\( x \)[/tex] represents the slope of the tangent line at any point [tex]\( x \)[/tex] on the curve. This can be done using the quotient rule:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}\left( \frac{8x}{x^2 + 1} \right) \][/tex]
Using the quotient rule, which is [tex]\(\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\)[/tex], we set:
[tex]\[ f(x) = 8x \quad \text{and} \quad g(x) = x^2 + 1 \][/tex]
Then, the derivatives are [tex]\( f'(x) = 8 \)[/tex] and [tex]\( g'(x) = 2x \)[/tex].

Applying the quotient rule:
[tex]\[ \frac{dy}{dx} = \frac{8(x^2 + 1) - 8x \cdot 2x}{(x^2 + 1)^2} = \frac{8x^2 + 8 - 16x^2}{(x^2 + 1)^2} = \frac{-8x^2 + 8}{(x^2 + 1)^2} = \frac{8(1 - x^2)}{(x^2 + 1)^2} \][/tex]

3. Evaluate the derivative at the point of tangency [tex]\( (1, 4) \)[/tex]:

We substitute [tex]\( x = 1 \)[/tex] into the derivative to find the slope of the tangent line at this point:
[tex]\[ \left. \frac{dy}{dx} \right|_{x=1} = \frac{8(1 - 1^2)}{(1^2 + 1)^2} = \frac{8(1 - 1)}{(1 + 1)^2} = \frac{8 \times 0}{4} = 0 \][/tex]
So, the slope of the tangent line at [tex]\( (1, 4) \)[/tex] is [tex]\( 0 \)[/tex].

4. Use the point-slope form to find the equation of the tangent line:

The point-slope form of a line is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\( m \)[/tex] is the slope, and [tex]\( (x_1, y_1) \)[/tex] is the point of tangency.
Substituting [tex]\( m = 0 \)[/tex], [tex]\( x_1 = 1 \)[/tex], and [tex]\( y_1 = 4 \)[/tex]:
[tex]\[ y - 4 = 0(x - 1) \][/tex]
Simplifying this, we obtain:
[tex]\[ y - 4 = 0 \implies y = 4 \][/tex]

So, the equation of the tangent line to the curve [tex]\( y = \frac{8x}{x^2 + 1} \)[/tex] at the point [tex]\((1, 4)\)[/tex] is [tex]\( y = 4 \)[/tex].