Question

Find the slope of tangent to
Y =
$4 \div x + 2 \sqrt{x}$
at x=4
​

Answer 1

Answer:

slope = $\frac{1}{4}$

Step-by-step explanation:

To find the slope, differentiate with respect to x and evaluate at x = 4

$\frac{dy}{dx}$ is the measure of the slope of the tangent at x = a

Differentiate each term using the power rule

$\frac{d}{dx}$ (a$x^{n}$ ) = na$x^{n-1}$

Given

y = $\frac{4}{x}$ + 2$\sqrt{x}$

  = 4$x^{-1}$ + 2$x^{\frac{1}{2} }$, hence

$\frac{dy}{dx}$ = - 4$x^{-2}$ + $x^{-\frac{1}{2} }$

$\frac{dy}{dx}$ = - $\frac{4}{x^{2} }$ + $\frac{1}{\sqrt{x} }$

At x = 4

$\frac{dy}{dx}$ = - $\frac{4}{16}$ + $\frac{1}{\sqrt{4} }$

                                   = - $\frac{1}{4}$ + $\frac{1}{2}$ = $\frac{1}{4}$