This article aims to find the differential of a given equation and the value of differential for given values of other parameters. Readers should know about differential equations and their basics to solve problems like in this article.
A differential equation is defined as an equation contains one or more terms and the derivatives of one variable (i.e., the dependent variable) concerning another variable (i.e., the independent variable)
\[\dfrac{dy}{dx} = f(x)\]
$x$ reprsents an independent variable, and $y$ is dependent variable.
Expert Answer
Given
\[ y = \sqrt { 15 + x ^ { 2 } } \]
The differential of $y$ is the derivative of a function times the differential of $ x $.
Therefore,
\[ dy = \dfrac { 1 } { 2 \sqrt { 15 + x ^ { 2 } } } . \dfrac { d } { dx } ( 15 + x ^ { 2 } ) . dx \]
\[\Rightarrow dy = \dfrac{1}{2 \sqrt {15+x^{2}}}.(0+2x)dx\]
\[dy = \dfrac{x}{\sqrt {15+x^{2}}} dx \]
Part (b)
Substituting $ x= 1 $ and $ dx = -0.2 $ in $ dy $ , we get
\[ \Rightarrow dy = \dfrac { 1 } { 15 + ( 1 ) ^ { 2 } } ( – 0.2 ) \]
\[ \Rightarrow dy = \dfrac { 1 } { \sqrt { 16 } } (- 0.2 ) \]
\[ \Rightarrow dy = \dfrac { – 0.2 } { 4 } \]
\[ \Rightarrow dy = – 0.05 \]
The value of $ dy $ for $ x= 1 $ and $ dx = -0.2 $ is $-0.05$
Numerical Result
– The differential $ dy $ is given as:
\[ dy = \dfrac { x } { \sqrt { 15 + x ^ { 2 }}} dx \]
– The value of $ dy $ for $ x= 1 $ and $ dx = -0.2 $ is $-0.05$
Example
(a) Find the differential $ dy $ for $ y = \sqrt { 20 – x ^ { 3 }} $.
(b) Evaluate $ dy $ for given values of $ x $ and $ dx $. $ x = 2 $, $ dx = – 0.2 $.
Solution
Given
\[ y = \sqrt { 20 – x ^ { 3 } } \]
The differential of $y$ is the derivative of a function times the differential of $ x $.
Therefore,
\[ dy = \dfrac {1} {2\sqrt { 20 – x^{3}}}.\dfrac { d } { dx } (20-x^{3}).dx \]
\[\Rightarrow dy = \dfrac{1}{2 \sqrt {20-x^{3}}}.(0-3x^{2})dx\]
\[dy = \dfrac{-3x^{2}}{2\sqrt {20-x^{3}}} dx \]
Part (b)
Substituting $x= 2$ and $dx = -0.2 $ in $dy$ , we get
\[ \Rightarrow dy = \dfrac {-3( 2 ) ^ { 2 } } { 2\sqrt {20 – (2) ^ { 3 }}} (- 0.2) \]
\[ \Rightarrow dy = \dfrac { -12 } { 4\sqrt { 3 }}(- 0.2)\]
\[ \Rightarrow dy = \dfrac { 2.4 } { 4 \sqrt { 3 } } \]
\[ \Rightarrow dy = 0.346 \]
The value of $ dy $ for $ x= 2 $ and $ dx = -0.2 $ is $0.346$