Question

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

Answer 1

Hint: In the given question, there are two equations. We will find the derivative of the first equation \[y = \dfrac{1}{4}{u^4}\] with respect to \[u\]. Then we will find the derivative of the second equation \[u = \dfrac{2}{3}{x^3} + 5\] with respect to \[x\]. By using these two derivatives, we will find \[\dfrac{{dy}}{{dx}}\].

Formula used
Power rule: \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]
\[\dfrac{{dc}}{{dx}} = 0\] where \[c\] is a constant
\[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]
\[\dfrac{d}{{dx}}\left( {cf\left( x \right)} \right) = c\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)\] where \[c\] is a constant
Sum rule: \[\dfrac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) + \dfrac{d}{{dx}}\left( {g\left( x \right)} \right)\]

Complete step by step solution:
Given equation is \[y = \dfrac{1}{4}{u^4}\]
Let’s calculate the derivate of the above equation with respect to \[u\].
\[\dfrac{{dy}}{{du}} = \dfrac{d}{{du}}\left( {\dfrac{1}{4}{u^4}} \right)\]
Now we will apply the formula \[\dfrac{d}{{dx}}\left( {cf\left( x \right)} \right) = c\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)\]
\[\dfrac{{dy}}{{du}} = \dfrac{1}{4}\dfrac{d}{{du}}\left( {{u^4}} \right)\]
Now we will apply the power rule of derivative
\[\dfrac{{dy}}{{du}} = \dfrac{1}{4} \times 4{u^{4 - 1}}\]
\[ \Rightarrow \dfrac{{dy}}{{du}} = {u^3}\]
The second equation is
\[u = \dfrac{2}{3}{x^3} + 5\]
Let’s calculate the derivate of the above equation with respect to \[x\].
\[\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{2}{3}{x^3} + 5} \right)\]
Now we will apply the sum rule of derivative.
\[\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{2}{3}{x^3}} \right) + \dfrac{d}{{dx}}\left( 5 \right)\]
Now we will apply the formulas \[\dfrac{d}{{dx}}\left( {cf\left( x \right)} \right) = c\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)\] and \[\dfrac{{dc}}{{dx}} = 0\]
\[\dfrac{{du}}{{dx}} = \dfrac{2}{3}\dfrac{d}{{dx}}\left( {{x^3}} \right) + 0\] [since 5 is a constant so it’s derivative will be zero]
Apply the power rule of derivative
\[\dfrac{{du}}{{dx}} = \dfrac{2}{3} \times 3{x^{3 - 1}}\]
\[ \Rightarrow \dfrac{{du}}{{dx}} = 2{x^2}\]
Let’s find the expression for \[\dfrac{{dy}}{{dx}}\] by applying the formula \[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]
Putting the values of \[\dfrac{{dy}}{{du}}\] and \[\dfrac{{du}}{{dx}}\]
\[\dfrac{{dy}}{{dx}} = 2{x^2} \cdot {u^3}\]
Now put \[u = \dfrac{2}{3}{x^3} + 5\] in the above equation
\[\dfrac{{dy}}{{dx}} = 2{x^2}{\left( {\dfrac{2}{3}{x^3} + 5} \right)^3}\]
Simplify the above equation
\[\dfrac{{dy}}{{dx}} = 2{x^2}{\left( {\dfrac{{2{x^3} + 5}}{3}} \right)^3}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2{x^2}}}{{27}}{\left( {2{x^3} + 5} \right)^3}\]

Hence the correct option is option C.

Note: Students are often confused with the formula \[\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{du}}}}{{\dfrac{{du}}{{dx}}}}\] and \[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]. But the correct formula is \[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]. \[\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{du}}}}{{\dfrac{{du}}{{dx}}}}\] is an incorrect formula. The correct formula should be \[\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{du}}}}{{\dfrac{{dx}}{{du}}}}\].