Question

If \[y = {t^{\dfrac{4}{3}}} - 3{t^{\dfrac{{ - 2}}{3}}}\], then what is the value of \[\dfrac{{dy}}{{dt}}\]?
A.\[\dfrac{{\left( {2{t^2} + 3} \right)}}{{2{t^{\dfrac{5}{3}}}}}\]
B.\[\dfrac{{\left( {2{t^2} + 3} \right)}}{{{t^{\dfrac{5}{3}}}}}\]
C.\[\dfrac{{2\left( {2{t^2} + 3} \right)}}{{{t^{\dfrac{5}{3}}}}}\]
D.\[\dfrac{{2\left( {2{t^2} + 3} \right)}}{{3{t^{\dfrac{5}{3}}}}}\]

Answer 1

Hint: In solving the above question, first differentiate each term with respect to \[t\], using derivative formula \[\dfrac{d}{{dx}}{x^n} = n \cdot {x^{\left( {n - 1} \right)}}\], and after that by simplifying the result we will get the desired result.

Formula used :

We will use the derivative formulas , \[\dfrac{d}{{dx}}{x^n} = n \cdot {x^{\left( {n - 1} \right)}}\], and exponential formula, \[{a^{ - m}} = \dfrac{1}{{{a^m}}}\].
Complete step by step solution:

Given \[y = {t^{\dfrac{4}{3}}} - 3{t^{\dfrac{{ - 2}}{3}}}\]
Now we will differentiate with respect to \[t\]on both sides,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left( {{t^{\dfrac{4}{3}}} - 3{t^{\dfrac{{ - 2}}{3}}}} \right)\],
Now we will distribute the differentiation on the right hand side, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}{t^{\dfrac{4}{3}}} - \dfrac{d}{{dt}}3{t^{\dfrac{{ - 2}}{3}}}\]
We will take out the constant term from the second term on the right hand side of the equation, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}{t^{\dfrac{4}{3}}} - 3\dfrac{d}{{dt}}{t^{\dfrac{{ - 2}}{3}}}\]
Now we will apply derivative formulas i.e., \[\dfrac{d}{{dx}}{x^n} = n \cdot {x^{\left( {n - 1} \right)}}\], then we will get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{4}{3}\left( {{t^{\dfrac{4}{3} - 1}}} \right) - 3\left( {\dfrac{{ - 2}}{3}} \right)\left( {{t^{\dfrac{{ - 2}}{3} - 1}}} \right)\]
Now we will simplify the expression we will get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{4}{3}\left( {{t^{\dfrac{{4 - 3}}{3}}}} \right) + 2\left( {{t^{\dfrac{{ - 2 - 3}}{3}}}} \right)\]
Now we will further simplify, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{4}{3}\left( {{t^{\dfrac{1}{3}}}} \right) + 2\left( {{t^{\dfrac{{ - 5}}{3}}}} \right)\]
Now we will take the L.C.M of the denominators i.e., L.C.M of 3 and 1 is 3, then we will get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{4{t^{\dfrac{1}{3}}} + 6{t^{\dfrac{{ - 5}}{3}}}}}{3}\]
Now we will take out the common factors, then we will get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{2\left( {2{t^{\dfrac{1}{3}}} + 3{t^{\dfrac{{ - 5}}{3}}}} \right)}}{3}\]
Now we will simplify further by taking out common factors, then we will get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{2{t^{\dfrac{{ - 5}}{3}}}\left( {2{t^2} + 3} \right)}}{3}\]
Now we will use exponential rule i.e, \[{a^{ - m}} = \dfrac{1}{{{a^m}}}\], then we will get,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{2\left( {2{t^2} + 3} \right)}}{{3{t^{\dfrac{5}{3}}}}}\].
The correct option is D.

Note: Students often do a common mistake when they simplify the indices. Sometimes they do \[{t^{ - \dfrac{5}{3}}} = \dfrac{1}{{{t^{\dfrac{3}{5}}}}}\] which is wrong. The correct one is \[{t^{ - \dfrac{5}{3}}} = \dfrac{1}{{{t^{\dfrac{5}{3}}}}}\].