Answer
a) $Δx = -0.5$
$dy = 0.5$
$Δy = 1$
b) $Δx = \frac{\pi}{4}$
$dy = \frac{\pi}{2}$
$Δy = 1$
c) $Δx = 3$
$dy = 0$
$Δy = -1$
Work Step by Step
a) As we have been given that x decreases from 2 to 1.5 so
$Δx = 1.5-2$
$Δx = -0.5$
As $y = \frac{1}{x-1}$
Differentiating :
$\frac{dy}{dx} = -\frac{1}{(x-1)^{2}}$
$dy = -\frac{1}{(x-1)^{2}}dx$
As $dx = Δx = -0.5$ and $x = 2$
$dy = -\frac{1}{(2-1)^{2}}(-0.5)$
$dy = 0.5$
and $Δy = \frac{1}{(1.5-1)}-\frac{1}{(2-1)}$
$Δy = 2-1$
$Δy = 1$
b) As we have been given that x decreases from $-\frac{\pi}{4}$ to $0$ so
$Δx = 0-(-\frac{\pi}{4})$
$Δx= \frac{\pi}{4}$
As $y = tanx$
Differentiating :
$\frac{dy}{dx} = sec^{2}x$
$dy = sec^{2}xdx$
As $dx = Δx = \frac{\pi}{4}$ and $x = -\frac{\pi}{4}$
$dy = sec^{2}(-\frac{\pi}{4})(\frac{\pi}{4})$
$dy = \frac{\pi}{2}$
And $Δy = tan0 - tan(-\frac{\pi}{4})$
$Δy = 1$
c) As we have been given that x decreases from 0 to 3 so
$Δx = 3-0$
$Δx = 3$
As $y =\sqrt( 25-x^{2})$
Differentiating :
$\frac{dy}{dx} = \frac{-x}{\sqrt (25-x^{2})}$
$dy = \frac{-x}{\sqrt (25-x^{2})}dx$
As $dx = Δx = 3$ and $x = 0$
$dy = \frac{-0}{\sqrt (25-(0)^{2})}(3)$
$dy = 0$
And $Δy = \sqrt (25-(3)^{2}) - \sqrt (25-(0)^{2})$
$Δy = 4-5$
$Δy = -1$
