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How do you use the chain rule to differentiate $ y=\dfrac{{{\left( {{x}^{3}}+4 \right)}^{5}}}{3{{x}^{4}}-2}$ ?

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Last updated date: 20th Jun 2024
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Answer
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Hint: In order to do this question, first you need to use the quotient rule to the differentiation. Quotient rule states that, if you have any equation in the form of $ y=\dfrac{u}{v}$, then you can differentiate the equation, by using the formula $ y'=\dfrac{u'v-v'u}{{{v}^{2}}}$. Here we get $ u={{\left( {{x}^{3}}+4 \right)}^{5}}$ and $ v=3{{x}^{4}}-2$. Then you can find the u’ and v’ and substitute all the required equations in the formula to get the final answer.

Complete step by step solution:
In order to do this question, first you need to use the quotient rule to the differentiation. Quotient rule states that, if you have any equation in the form of $ y=\dfrac{u}{v}$, then you can differentiate the equation, by using the formula $ y'=\dfrac{u'v-v'u}{{{v}^{2}}}$. Here we get $ u={{\left( {{x}^{3}}+4 \right)}^{5}}$ and $ v=3{{x}^{4}}-2$. First we need to find the values for u’ and v’.
To find u’ we use chain rule. Therefore, we get the u’ as
$ \Rightarrow u={{\left( {{x}^{3}}+4 \right)}^{5}}$
$ \Rightarrow u'=5{{\left( {{x}^{3}}+4 \right)}^{4}}\dfrac{d}{dx}\left( {{x}^{3}}+4 \right)$
$ \Rightarrow u'=5{{\left( {{x}^{3}}+4 \right)}^{4}}\left( 3{{x}^{2}} \right)$
Now we have to find v’
$ \Rightarrow v=3{{x}^{4}}-2$
$ \Rightarrow v=12{{x}^{3}}$
Therefore, now we have to substitute the value of u, v, u’, v’ in the quotient rule formula to get the final answer. Finally, we get
$ \Rightarrow y'=\dfrac{u'v-v'u}{{{v}^{2}}}$
$ \Rightarrow y'=\dfrac{5{{\left( {{x}^{3}}+4 \right)}^{4}}\left( 3{{x}^{2}} \right)\left( 3{{x}^{4}}-2 \right)-12{{x}^{3}}\left( {{\left( {{x}^{3}}+4 \right)}^{5}} \right)}{{{\left( 3{{x}^{4}}-2 \right)}^{2}}}$
$ \Rightarrow y'=\dfrac{{{\left( {{x}^{3}}+4 \right)}^{4}}\left( \left( 45{{x}^{6}}-30{{x}^{2}} \right)-12{{x}^{3}}\left( {{x}^{3}}+4 \right) \right)}{{{\left( 3{{x}^{4}}-2 \right)}^{2}}}$
$ \Rightarrow y'=\dfrac{{{\left( {{x}^{3}}+4 \right)}^{4}}\left( 45{{x}^{6}}-30{{x}^{2}}-12{{x}^{6}}-48{{x}^{3}} \right)}{{{\left( 3{{x}^{4}}-2 \right)}^{2}}}$
$ \Rightarrow y'=\dfrac{{{\left( {{x}^{3}}+4 \right)}^{4}}\left( 33{{x}^{6}}-30{{x}^{2}}-48{{x}^{3}} \right)}{{{\left( 3{{x}^{4}}-2 \right)}^{2}}}$
Therefore, we get the final answer for the differentiation as $ y'=\dfrac{{{\left( {{x}^{3}}+4 \right)}^{4}}\left( 33{{x}^{6}}-30{{x}^{2}}-48{{x}^{3}} \right)}{{{\left( 3{{x}^{4}}-2 \right)}^{2}}}$.

Note: To do this question, you need to know the basic rules and formulas of the differentiation like the product rule, chain rule, quotient rule, and others. Also you need to know the basics of differentiation. They will really help to solve the questions easily.