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I'm going to assume you mean
[tex]y = 14t^2 - \dfrac{t^2}{2+5t} - 3[/tex]
Taking the derivative with respect to t on both sides gives
[tex]\dfrac{\mathrm dy}{\mathrm dt} = 2\cdot14t - \dfrac{(2+5t)(2t)-t^2(5)}{(2+5t)^2} - 0 \\\\ \dfrac{\mathrm dy}{\mathrm dt} = \boxed{28t - \dfrac{4t+5t^2}{(2+5t)^2}}[/tex]
You could go on to expand the rational expression into partial fractions. One way to do this:
[tex]\dfrac{5t^2 + 4t}{(2 + 5t)^2} = \dfrac{t(5t+2) + 2t}{(2+5t)^2} \\\\= \dfrac{t}{2+5t} + \dfrac{2t}{(2+5t)^2} \\\\= \dfrac15 \cdot \dfrac{t}{\frac25+t} + \dfrac{2t}{(2+5t)^2} \\\\ = \dfrac15 \cdot \dfrac{t+\frac25-\frac25}{\frac25+t} + \dfrac{2t}{(2+5t)^2} \\\\ = \dfrac15 \left(1 - \dfrac{2}{2+5t}\right) + \dfrac{2t}{(2+5t)^2} \\\\ = \dfrac15 - \dfrac{2}{10+25t}+ \dfrac{2t}{(2+5t)^2}[/tex]
and this would give
[tex]\dfrac{\mathrm dy}{\mathrm dt} = 28t - \dfrac15 + \dfrac2{10+25t} - \dfrac{2t}{(2+5t)^2}[/tex]