Questions: Find an equation of the tangent line to the curve at the given point. y=5x-4√x,(1,1) y=

Solution Steps

To find the equation of the tangent line, we first need to find the derivative of the function $y = 5x - 4\sqrt{x}$. The derivative, $y'$, will give us the slope of the tangent line at any point $x$.

The derivative of $5x$ is $5$.

The derivative of $-4\sqrt{x}$ is found using the power rule. Rewrite $\sqrt{x}$ as $x^{1/2}$, so the derivative is:

$$\frac{d}{dx}(-4x^{1/2}) = -4 \cdot \frac{1}{2}x^{-1/2} = -2x^{-1/2} = -\frac{2}{\sqrt{x}}$$

Thus, the derivative of the function is:

$$y' = 5 - \frac{2}{\sqrt{x}}$$

We need to find the slope of the tangent line at the point $(1, 1)$. Substitute $x = 1$ into the derivative:

$$y'(1) = 5 - \frac{2}{\sqrt{1}} = 5 - 2 = 3$$

The slope of the tangent line at $(1, 1)$ is $3$.

The point-slope form of a line is given by:

$$y - y_1 = m(x - x_1)$$

where $m$ is the slope and $(x_1, y_1)$ is the point on the line. Here, $m = 3$ and $(x_1, y_1) = (1, 1)$.

Substitute these values into the point-slope form:

$$y - 1 = 3(x - 1)$$

Simplify the equation:

$$y - 1 = 3x - 3$$

$$y = 3x - 2$$

Final Answer