Question

11-18: Find the differential of the function.

12. \(y = \sqrt {1 - {t^4}} \)

Short answer

The differential of the function is \(dy = - \frac{{2{t^3}}}{{\sqrt {1 - {t^4}} }}dt\).

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

The function is \(y = f\left( t \right) = \sqrt {1 - {t^4}} \).

Evaluate the derivative of the function as shown below:

\(\begin{aligned}f'\left( t \right) &= \frac{d}{{dt}}\left( {\sqrt {1 - {t^4}} } \right)\\ &= \frac{d}{{dt}}\left( {{{\left( {1 - {t^4}} \right)}^{\frac{1}{2}}}} \right)\\ &= \frac{1}{2}{\left( {1 - {t^4}} \right)^{ - \frac{1}{2}}}\left( { - 4{t^3}} \right)\\ &= - \frac{{2{t^3}}}{{\sqrt {1 - {t^4}} }}\end{aligned}\)

Obtain the differential of the function as shown below:

\(\begin{aligned}dy &= f'\left( t \right)dt\\dy &= - \frac{{2{t^3}}}{{\sqrt {1 - {t^4}} }}dt\end{aligned}\)

Thus, the differential of the function is \(dy = - \frac{{2{t^3}}}{{\sqrt {1 - {t^4}} }}dt\).