Question

If \[y = (\dfrac{1}{4}){u^4},u = (\dfrac{2}{3}){x^3} + 5\], then \[\dfrac{{dy}}{{dx}} = \]
A. \[(\dfrac{{{x^2}}}{{27}}){(2{x^3} + 15)^3}\]
B. \[(\dfrac{{2x}}{{27}}){(2{x^3} + 15)^3}\]
C. \[(\dfrac{{2{x^2}}}{{27}}){(2{x^3} + 15)^3}\]
D. None of these

Answer 1

Hint: This question involves the basic concepts of differentiation. Then the concept of differentiation named differentiation of parametric function is used here.According to it, instead of a function y(x) being defined explicitly in terms of the independent variable x, it is sometimes helpful to define both x and y in terms of a third variable, t say, known as a parameter.

Formula used:
Differentiation is given by:
\[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}}\]
where $y$ and $x$ are both a function of another parameter $t$.

Complete step by step answer:
Let us begin the question with the given equations, which are
\[y = (\dfrac{1}{4}){u^4}\]and \[u = (\dfrac{2}{3}){x^3} + 5\]
Now, by using the concept of differentiation of parametric functions, we can write \[\dfrac{{dy}}{{dx}}\] as
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}\]
Now, let us solve \[\dfrac{{dy}}{{du}}\] and \[\dfrac{{du}}{{dx}}\] separately,
Firstly, let's start solving \[\dfrac{{dy}}{{du}}\]
\[ \Rightarrow \dfrac{{dy}}{{du}} = \dfrac{{d(\dfrac{1}{4}){u^4}}}{{du}}\]
Now, by applying the formula, \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\] we get,
\[ \Rightarrow \dfrac{{dy}}{{du}} = (\dfrac{1}{4}) \times 4{u^3}\]

Simplifying the right-hand side
\[ \Rightarrow \dfrac{{dy}}{{du}} = {u^3} - - - - (i)\]
Now, let us start solving \[\dfrac{{du}}{{dx}}\]
\[ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{d((\dfrac{2}{3}){x^3} + 5)}}{{dx}}\]
Now, by using the formula \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]we get,
\[ \Rightarrow \dfrac{{du}}{{dx}} = (\dfrac{2}{3}) \times 3{x^2}\]
Simplifying the right-hand side
\[ \Rightarrow \dfrac{{du}}{{dx}} = 2{x^2} - - - - (ii)\]
Now, by using the equations\[(i)\]and \[(ii)\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {u^3} \times 2{x^2}\],

Now, by replacing the value of u in terms of x we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {u^3} \times 2{x^2}\]
Now, by putting \[u = (\dfrac{2}{3}){x^3} + 5\]we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {((\dfrac{2}{3}){x^3} + 5)^3} \times 2{x^2}\]
Taking three as the common denominator
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {(\dfrac{{2{x^3} + 15}}{3})^3} \times 2{x^2}\]
Now, taking 3 out of the bracket
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{(2{x^3} + 15)}^3}}}{{27}} \times 2{x^2}\]
And by grouping 27 with 2x2 we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{(2{x^3} + 15)}^3}}}{{27}} \times 2{x^2}\]
And finally, we arrive at the answer
\[ \therefore \dfrac{{dy}}{{dx}} = (\dfrac{{2{x^2}}}{{27}}){(2{x^3} + 15)^3}\]

Therefore, option C is the correct answer.

Note:This question is based on the concept of differentiation of parametric equations. One should be well versed with this concept before solving the question. Do not commit calculation mistakes, and be sure of the final answer.In calculus, differentiation is one of the two important concepts apart from integration. Differentiation is a method of finding the derivative of a function. Differentiation is a process, in Maths, where we find the instantaneous rate of change in function based on one of its variables.