Question

19-22: (a) Find the differential \(dy\) and (b) evaluate \(dy\) for the given values of x and \(dx\).

21. \(y = \sqrt {3 + {x^2}} ,{\rm{ }}x = 1,{\rm{ }}dx = - 0.1\)

Short answer

a) The differential of the function is \(dy = \frac{x}{{\sqrt {3 + {x^2}} }}dx\).

b) The value of \(dy\) is \( - 0.05\).

Step-by-step solution

Definition of differentials

The equation establishes the differential \(dy\)with respect to \(dx\) as shown below:

\(dy = f'\left( x \right)dx\)

Where \(dx\) represent the independent variableand \(dy\) represent the dependent variable.

Determine the differential of the function

a)

The function is \(y = f\left( x \right) = \sqrt {3 + {x^2}} \).

Obtain the differential of the function as shown below:

\(\begin{aligned}dy &= f'\left( x \right)dx\\dy &= \frac{1}{2}{\left( {3 + {x^2}} \right)^{ - \frac{1}{2}}}\left( {2x} \right)dx\\dy &= \frac{{2x}}{{2\sqrt {3 + {x^2}} }}dx\\dy &= \frac{x}{{\sqrt {3 + {x^2}} }}dx\end{aligned}\)

Thus, the differential of the function is \(dy = \frac{x}{{\sqrt {3 + {x^2}} }}dx\).

Evaluate the value of \(dy\)

b)

Substitute\(x = 1\)and\(dx = - 0.1\)to obtain\(dy\)as shown below:

\(\begin{aligned}dy &= \frac{1}{{\sqrt {3 + {1^2}} }}\left( { - 0.1} \right)\\ &= \frac{1}{{\sqrt 4 }}\left( { - 0.1} \right)\\ &= \frac{1}{2}\left( { - 0.1} \right)\\ &= - 0.05\end{aligned}\)

Thus, the value of \(dy\) is \( - 0.05\).