(a) Find the differential dy. y=24+x2 (b)Evaluate dy for the given values of x and dx. x=1, dx=−0.2

Find the derivative using chain rule. Then multiply both sides by dx. (a)y=24+x2 y=(24+x2)12 dydx=12(24+x2)−12(2x) =x24+x2 dy=x24+x2dx Substitute x=1 and dx=−0.2 into the differential from part (a). Then evaluate. (b)dy=124+12(−0.2) =125(−0.2) =15(−0.2) =0.2(-0.2) =-0.04 Results: (a)dy=x24+x2dx (b)dy=−0.04

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College
Calculus and Analysis
Differential Calculus
Derivatives

waigaK

Answered question

2021-10-19

(a) Find the differential dy. $y = \sqrt{24 + x^{2}}$
(b)Evaluate dy for the given values of x and dx.
$x = 1, d x = - 0.2$

Answer & Explanation

cheekabooy

Skilled2021-10-20Added 83 answers

Find the derivative using chain rule. Then multiply both sides by dx.
(a) $y = \sqrt{24 + x^{2}}$
$y = {(24 + x^{2})}^{\frac{1}{2}}$
$\frac{d y}{d x} = \frac{1}{2} {(24 + x^{2})}^{- \frac{1}{2}} (2 x)$
$= \frac{x}{\sqrt{24 + x^{2}}}$
$d y = \frac{x}{\sqrt{24 + x^{2}}} d x$
Substitute $x = 1$ and $d x = - 0.2$ into the differential from part (a). Then evaluate.
(b) $d y = \frac{1}{\sqrt{24 + 1^{2}}} (- 0.2)$
$= \frac{1}{\sqrt{25}} (- 0.2)$
$= \frac{1}{5} (- 0.2)$
=0.2(-0.2)
=-0.04
Results:
(a) $d y = \frac{x}{\sqrt{24 + x^{2}}} d x$
(b) $d y = - 0.04$

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