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Question: How do you find the derivative of \[f\left( x \right)=\left( x-1 \right){{\left( {{x}^{2}}+2 \right)...

How do you find the derivative of f(x)=(x−1)(x2+2)3f\left( x \right)=\left( x-1 \right){{\left( {{x}^{2}}+2 \right)}^{3}}?

Explanation

Solution

From the question we have been asked to find the derivative of f(x)=(x−1)(x2+2)3f\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(x)=(x−1)(x2+2)3f\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(x)h(x)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: dydx=g(x)h′(x)+h(x)g′(x)\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(x)=x−1g\left( x \right)=x-1 and h(x)=(x2+2)3h\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′(x)=1{g}'\left( x \right)=1 and
h′(x)=3(x2+2)2.ddx(x2+2){h}'\left( x \right)=3{{\left( {{x}^{2}}+2 \right)}^{2}}.\dfrac{d}{dx}\left( {{x}^{2}}+2 \right)
⇒h′(x)=6x(x2+2)2\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
dydx=g(x)h′(x)+h(x)g′(x)\dfrac{dy}{dx}=g\left( x \right){h}'\left( x \right)+h\left( x \right){g}'\left( x \right)
⇒dydx=6x(x−1)(x2+2)2+(x2+2)3\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
dydx=(x2+2)2(6x2−6x+x2+2)\dfrac{dy}{dx}={{\left( {{x}^{2}}+2 \right)}^{2}}\left( 6{{x}^{2}}-6x+{{x}^{2}}+2 \right)
⇒dydx=(x2+2)2(7x2−6x+2)\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 ddx(uv)=v(dudx)−u(dvdx)v2\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.