Question

3-34:Differentiate the function.

31. \(D\left( t \right) = \frac{{1 + 16{t^2}}}{{{{\left( {4t} \right)}^3}}}\)

Short answer

The differentiation of the function is\(\frac{{dD}}{{dt}} = - \frac{{3 + 16{t^2}}}{{64{t^4}}}\).

Step-by-step solution

Differentiation of the Function

When we differentiate any function with respect to any variable present in that function, then basically we are determining the slope of that function that ranges within the domain of that variable.

Different rules of differentiation

The given function is \(D\left( t \right) = \frac{{1 + 16{t^2}}}{{{{\left( {4t} \right)}^3}}}\).

The Quotient rule can be explained using the general form as: \(\frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{\left( {g\left( x \right)} \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right) - \left( {f\left( x \right)} \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{\left( {g\left( x \right)} \right)}}\).

The sum rule is:\(\frac{d}{{dx}}\left( {g\left( x \right) + f\left( x \right)} \right) = \frac{d}{{dx}}\left( {g\left( x \right)} \right) + \frac{d}{{dx}}\left( {f\left( x \right)} \right)\)

And, the power rule is: \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\)

Evaluating Derivative using all these rules

Using these rules, differentiating the given function with respect to\(t\)as:

\(\begin{align}D'\left( t \right) &= \frac{d}{{dt}}\left( {\frac{{1 + 16{t^2}}}{{{{\left( {4t} \right)}^3}}}} \right)\\ &= \frac{{{{\left( {4t} \right)}^3}\frac{d}{{dt}}\left( {1 + 16{t^2}} \right) - \left( {1 + 16{t^2}} \right)\frac{d}{{dt}}{{\left( {4t} \right)}^3}}}{{{{\left\{ {{{\left( {4t} \right)}^3}} \right\}}^2}}}\\ &= \frac{{{{\left( {4t} \right)}^3}\left\{ {\frac{d}{{dt}}\left( 1 \right) + \frac{d}{{dt}}\left( {16{t^2}} \right)} \right\} - \left( {1 + 16{t^2}} \right)\frac{d}{{dt}}{{\left( {4t} \right)}^3}}}{{{{\left\{ {{{\left( {4t} \right)}^3}} \right\}}^2}}}\\ &= \frac{{{{\left( {4t} \right)}^3} \cdot \left( {32t} \right) - \left( {1 + 16{t^2}} \right) \cdot \left( {64 \cdot \left( {3{t^2}} \right)} \right)}}{{{{\left( {4t} \right)}^6}}}\\ &= \frac{{64{t^2}\left\{ {32{t^2} - 3 - 48{t^2}} \right\}}}{{{{\left( {4t} \right)}^6}}}\\ &= - \frac{{3 + 16{t^2}}}{{64{t^4}}}\end{align}\)

Hence, the derivative of the given function is \(D'\left( t \right) = - \frac{{3 + 16{t^2}}}{{64{t^4}}}\).