This**Â article aims**Â to find the **differential of a given equation**Â and the value of **differential**Â for given values of other **parameters**. Readers should know about **differential equations**Â and their **basics to solve problems**Â like in this article.

A **differential equation**Â is defined as an equation contains one or more terms and the **derivatives of one variable**Â (i.e., the **dependent variable**) concerning another **variable**Â (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**Â differential**Â of $y$ is the**Â derivative of a function times**Â the differential of $ x $.

**Therefore,**

\[ dy = \dfrac { 1 } { 2 \sqrt { 15 + x ^ { 2 } } } . \dfrac { d } { dx } ( 15 + x ^ { 2 } ) . dx \]

\[\Rightarrow Â dy = \dfrac{1}{2 \sqrt {15+x^{2}}}.(0+2x)dx\]

\[dy = \dfrac{x}{\sqrt {15+x^{2}}} dx \]

**Part (b)**

**Substituting**Â $ 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**Â differential**Â of $y$ is the**Â derivative of a function times**Â the differential of $ x $.

**Therefore,**

\[ dy = \dfrac {1} {2\sqrt { 20 – x^{3}}}.\dfrac { d } { dx } (20-x^{3}).dx \]

\[\Rightarrow Â dy = \dfrac{1}{2 \sqrt {20-x^{3}}}.(0-3x^{2})dx\]

\[dy = \dfrac{-3x^{2}}{2\sqrt {20-x^{3}}} dx \]

**Part (b)**

**Substituting**Â $x= 2$ 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 $0.346$**