Question

If we have an expression as $y = {t^{\dfrac{4}{3}}} - 3{t^{\dfrac{{ - 2}}{3}}}$, then $\dfrac{{dy}}{{dt}} =$
$A)\dfrac{{2{t^2} + 3}}{{3{t^{\dfrac{5}{3}}}}}$
$B)\dfrac{{2{t^2} + 3}}{{{t^{\dfrac{5}{3}}}}}$
$C)\dfrac{{2(2{t^2} + 3)}}{{{t^{\dfrac{5}{3}}}}}$
$D)\dfrac{{2(2{t^2} + 3)}}{{3{t^{\dfrac{5}{3}}}}}$

Answer 1

First, we need to analyze the given information which is in the form of $y = {t^{\dfrac{4}{3}}} - 3{t^{\dfrac{{ - 2}}{3}}}$ and we need to find its derivative part.
In differentiation, the derivative of $x$ raised to the power is denoted by $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ .
By using the definition of differentiation, we will easily solve the given problem.
The inverse of the function is defined as ${x^{ - 1}} = \dfrac{1}{x}$

Complete step-by-step solution:
Since from the problem given that we have $y = {t^{\dfrac{4}{3}}} - 3{t^{\dfrac{{ - 2}}{3}}}$.
Now let us take ${t^{\dfrac{4}{3}}}$ and applying the differentiation formula, we get $\dfrac{{d({t^{\dfrac{4}{3}}})}}{{dt}} = \dfrac{4}{3} \times {t^{\dfrac{4}{3} - 1}}$ (by applying the formula of the general derivation)
Further solving we get $\dfrac{{d({t^{\dfrac{4}{3}}})}}{{dt}} = \dfrac{4}{3} \times {t^{\dfrac{4}{3} - 1}} \Rightarrow \dfrac{4}{3} \times {t^{\dfrac{1}{3}}}$
Now take the second value $- 3{t^{\dfrac{{ - 2}}{3}}}$ and applying the differentiation formula we get $\dfrac{{d( - 3{t^{\dfrac{{ - 2}}{3}}})}}{{dt}} = - 3( - \dfrac{2}{3}) \times {t^{\dfrac{{ - 2}}{3} - 1}}$ ((by applying the formula of the general derivation)
Further solving we get $\dfrac{{d( - 3{t^{\dfrac{{ - 2}}{3}}})}}{{dt}} = - 3( - \dfrac{2}{3}) \times {t^{\dfrac{{ - 2}}{3} - 1}} \Rightarrow 2 \times {t^{\dfrac{{ - 5}}{3}}}$
Hence in general differentiation we have $y = {t^{\dfrac{4}{3}}} - 3{t^{\dfrac{{ - 2}}{3}}}$, then we get $\dfrac{{dy}}{{dx}} = \dfrac{4}{3} \times {t^{\dfrac{1}{3}}} + 2 \times {t^{\dfrac{{ - 5}}{3}}}$ (where $\dfrac{{d({t^{\dfrac{4}{3}}})}}{{dt}} = \dfrac{4}{3} \times {t^{\dfrac{4}{3} - 1}} \Rightarrow \dfrac{4}{3} \times {t^{\dfrac{1}{3}}}$ and $\dfrac{{d( - 3{t^{\dfrac{{ - 2}}{3}}})}}{{dt}} = - 3( - \dfrac{2}{3}) \times {t^{\dfrac{{ - 2}}{3} - 1}}$)
Thus, solving the equation, we have $\dfrac{{dy}}{{dx}} = \dfrac{4}{3} \times {t^{\dfrac{1}{3}}} + 2 \times {t^{\dfrac{{ - 5}}{3}}} = \dfrac{4}{3} \times {t^{\dfrac{1}{3}}} + 2 \times \dfrac{1}{{{t^{\dfrac{5}{3}}}}}$ (by the inverse rule)
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{3} \times {t^{\dfrac{1}{3}}} + 2 \times \dfrac{1}{{{t^{\dfrac{5}{3}}}}}=\dfrac{{4{t^{\dfrac{1}{3}}}{t^{\dfrac{5}{3}}} + 2 \times 3}}{{3{t^{\dfrac{5}{3}}}}}$ (by the cross multiplication)
Hence, we get $\dfrac{{dy}}{{dx}} = \dfrac{{4{t^{\dfrac{1}{3}}}{t^{\dfrac{5}{3}}} + 2 \times 3}}{{3{t^{\dfrac{5}{3}}}}} = \dfrac{{2(2{t^{\dfrac{1}{3} + \dfrac{5}{3}}} + 3)}}{{3{t^{\dfrac{5}{3}}}}}$ (by the power rule ${x^1} \times {x^1} = {x^{1 + 1}} = {x^2}$)
Thus, we have $\dfrac{{dy}}{{dx}} = \dfrac{{2(2{t^{\dfrac{6}{3}}} + 3)}}{{3{t^{\dfrac{5}{3}}}}} \Rightarrow \dfrac{{2(2{t^2} + 3)}}{{3{t^{\dfrac{5}{3}}}}}$ (which is the required form of the derivation part)
Therefore option $D)\dfrac{{2(2{t^2} + 3)}}{{3{t^{\dfrac{5}{3}}}}}$ is correct.

Note: Differentiation can be defined as the derivative of the independent variable value and can be used to calculate features in an independence variance per unit modification.
In other words, derivative is actually the rate of change of a particular function.
So, the differentiation formula is $\dfrac{{dy}}{{dx}}$ . It shows that the difference in y is divided by the difference in x and also d is not the variable.
There are some differentiation rules, which are Sum and difference rule, Product rule, Quotient rule, and Chain rule.
Differentiation and integration are inverse processes like a derivative of $\dfrac{{d({x^2})}}{{dx}} = 2x$ and the integration is $\int {2xdx = \dfrac{{2{x^2}}}{2}} \Rightarrow {x^2}$