
How do you find the derivative of \[f\left( x \right)=\left( x-1 \right){{\left( {{x}^{2}}+2 \right)}^{3}}\]?
Answer
544.8k+ views
Hint: From the question we have been asked to find the derivative of \[f\left( x \right)=\left( x-1 \right){{\left( {{x}^{2}}+2 \right)}^{3}}\].We can find the derivative for the given question by using the basic formulae of differentiation. After applying the formulae, we have to do a lot more calculation to get the answer.
Complete answer:
From the question, we have been given that \[f\left( x \right)=\left( x-1 \right){{\left( {{x}^{2}}+2 \right)}^{3}}\]
First of all, we can clearly observe that it is in the form of \[y=g\left( x \right)h\left( x \right)\], that is it is in the form of a product of two functions.
For this type of questions, we have product rules in differentiation to differentiate the given question.
Product rule formula is shown below: \[\dfrac{dy}{dx}=g\left( x \right){h}'\left( x \right)+h\left( x \right){g}'\left( x \right)\]
Now, we have to apply the above product rule formula for our given question to find its derivative.
From the question given, we have been given that,
\[g\left( x \right)=x-1\] and \[h\left( x \right)={{\left( {{x}^{2}}+2 \right)}^{3}}\]
Now, we have to find derivatives of the above two functions to get them substituted in the product rule formula.
\[{g}'\left( x \right)=1\] and
\[{h}'\left( x \right)=3{{\left( {{x}^{2}}+2 \right)}^{2}}.\dfrac{d}{dx}\left( {{x}^{2}}+2 \right)\]
\[\Rightarrow {h}'\left( x \right)=6x{{\left( {{x}^{2}}+2 \right)}^{2}}\]
Now, we got all the values needed to substitute in the product rule formula.
So, substitute the all values we got in the product rule formula,
By substituting all the values we got in the product rule formula, we get
\[\dfrac{dy}{dx}=g\left( x \right){h}'\left( x \right)+h\left( x \right){g}'\left( x \right)\]
\[\Rightarrow \dfrac{dy}{dx}=6x\left( x-1 \right){{\left( {{x}^{2}}+2 \right)}^{2}}+{{\left( {{x}^{2}}+2 \right)}^{3}}\]
Now, by taking the common terms out and simplifying, we get
\[\dfrac{dy}{dx}={{\left( {{x}^{2}}+2 \right)}^{2}}\left( 6{{x}^{2}}-6x+{{x}^{2}}+2 \right)\]
\[\Rightarrow \dfrac{dy}{dx}={{\left( {{x}^{2}}+2 \right)}^{2}}\left( 7{{x}^{2}}-6x+2 \right)\]
Hence, the derivative for the given question is found.
Note: We should be very careful while finding the derivatives. We should be very careful while doing the calculation especially while doing the calculation using the product rule formula. Also, we should be well known about the basic formulae of differentiation and how to use them in the correct steps of the given question. Similarly as we have product rule we also have division rule known as \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\left( \dfrac{du}{dx} \right)-u\left( \dfrac{dv}{dx} \right)}{{{v}^{2}}}\] that we can use while solving questions of this type.
Complete answer:
From the question, we have been given that \[f\left( x \right)=\left( x-1 \right){{\left( {{x}^{2}}+2 \right)}^{3}}\]
First of all, we can clearly observe that it is in the form of \[y=g\left( x \right)h\left( x \right)\], that is it is in the form of a product of two functions.
For this type of questions, we have product rules in differentiation to differentiate the given question.
Product rule formula is shown below: \[\dfrac{dy}{dx}=g\left( x \right){h}'\left( x \right)+h\left( x \right){g}'\left( x \right)\]
Now, we have to apply the above product rule formula for our given question to find its derivative.
From the question given, we have been given that,
\[g\left( x \right)=x-1\] and \[h\left( x \right)={{\left( {{x}^{2}}+2 \right)}^{3}}\]
Now, we have to find derivatives of the above two functions to get them substituted in the product rule formula.
\[{g}'\left( x \right)=1\] and
\[{h}'\left( x \right)=3{{\left( {{x}^{2}}+2 \right)}^{2}}.\dfrac{d}{dx}\left( {{x}^{2}}+2 \right)\]
\[\Rightarrow {h}'\left( x \right)=6x{{\left( {{x}^{2}}+2 \right)}^{2}}\]
Now, we got all the values needed to substitute in the product rule formula.
So, substitute the all values we got in the product rule formula,
By substituting all the values we got in the product rule formula, we get
\[\dfrac{dy}{dx}=g\left( x \right){h}'\left( x \right)+h\left( x \right){g}'\left( x \right)\]
\[\Rightarrow \dfrac{dy}{dx}=6x\left( x-1 \right){{\left( {{x}^{2}}+2 \right)}^{2}}+{{\left( {{x}^{2}}+2 \right)}^{3}}\]
Now, by taking the common terms out and simplifying, we get
\[\dfrac{dy}{dx}={{\left( {{x}^{2}}+2 \right)}^{2}}\left( 6{{x}^{2}}-6x+{{x}^{2}}+2 \right)\]
\[\Rightarrow \dfrac{dy}{dx}={{\left( {{x}^{2}}+2 \right)}^{2}}\left( 7{{x}^{2}}-6x+2 \right)\]
Hence, the derivative for the given question is found.
Note: We should be very careful while finding the derivatives. We should be very careful while doing the calculation especially while doing the calculation using the product rule formula. Also, we should be well known about the basic formulae of differentiation and how to use them in the correct steps of the given question. Similarly as we have product rule we also have division rule known as \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\left( \dfrac{du}{dx} \right)-u\left( \dfrac{dv}{dx} \right)}{{{v}^{2}}}\] that we can use while solving questions of this type.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

