Question

Differentiate $\left( {{x}^{2}}-5x+8 \right)\left( {{x}^{3}}+7x+9 \right)$in three ways mentioned below,
(i) By using product rule
(ii) By expanding the product to obtain a single polynomial
(iii) By logarithmic differentiation
Do they all give the same answer?

Answer 1

To solve the question at first we consider the two polynomials given in the question as separate functions that means $f(x)=\left( {{x}^{2}}-5x+8 \right)$and$g(x)=\left( {{x}^{3}}+7x+9 \right)$.
Then we use the product rule of differentiation of two functions that is\dfrac{d}{dx}\left\\{ f(x)\cdot g(x) \right\\}=f(x)\dfrac{d}{dx}\left\\{ g(x) \right\\}+g(x)\dfrac{d}{dx}\left\\{ f(x) \right\\}. Then we have to expand the product of the given expression in terms of single polynomial and differentiate separate terms using the differentiation rule$\dfrac{d}{dx}\left( u+v+w+.......... \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}+\dfrac{dw}{dx}+........$. Finally consider the whole expression as $h(x)=\left( {{x}^{2}}-5x+8 \right)\left( {{x}^{3}}+7x+9 \right)$and we have to take logarithm with base e on both the sides and separating the RHS using logarithmic formula${{\log }_{e}}\left( ab \right)={{\log }_{e}}a+{{\log }_{e}}b$, differentiate it. Then we have to match the obtained answers to check if the answers are the same or not.

Complete step-by-step answer:
Let’s consider
$f(x)=\left( {{x}^{2}}-5x+8 \right)$………………………….. (1)
And $g(x)=\left( {{x}^{3}}+7x+9 \right)$…………………………. (2)
(i) Differentiation by using product rule
We know the product rule for the differentiation which is given by,
\dfrac{d}{dx}\left\\{ f(x)\cdot g(x) \right\\}=f(x)\dfrac{d}{dx}\left\\{ g(x) \right\\}+g(x)\dfrac{d}{dx}\left\\{ f(x) \right\\} ………………….. (3)
Then applying the rule we can write \dfrac{d}{dx}\left\\{ \left( {{x}^{2}}-5x+8 \right)\left( {{x}^{3}}+7x+9 \right) \right\\}=\left( {{x}^{2}}-5x+8 \right)\dfrac{d}{dx}\left( {{x}^{3}}+7x+9 \right)+\left( {{x}^{3}}+7x+9 \right)\dfrac{d}{dx}\left( {{x}^{2}}-5x+8 \right)
=\left( {{x}^{2}}-5x+8 \right)\left\\{ \dfrac{d}{dx}\left( {{x}^{3}} \right)+\dfrac{d}{dx}\left( 7x \right)+\dfrac{d}{dx}\left( 9 \right) \right\\}+\left( {{x}^{3}}+7x+9 \right)\left\\{ \dfrac{d}{dx}\left( {{x}^{2}} \right)-\dfrac{d}{dx}\left( 5x \right)+\dfrac{d}{dx}\left( 8 \right) \right\\}
$=\left( {{x}^{2}}-5x+8 \right)\left( 3{{x}^{2}}+7+0 \right)+\left( {{x}^{3}}+7x+9 \right)\left( 2x-5+0 \right)$
$=3{{x}^{4}}+7{{x}^{2}}-15{{x}^{3}}-35x+24{{x}^{2}}+56+2{{x}^{4}}-5{{x}^{3}}+14{{x}^{2}}-35x+18x-45$
$=5{{x}^{4}}-20{{x}^{3}}+45{{x}^{2}}-52x+9$
(ii) Differentiation by expanding the product to obtain a single polynomial
Now let’s find out the product$\left( {{x}^{2}}-5x+8 \right)\left( {{x}^{3}}+7x+9 \right)$
$\left( {{x}^{2}}-5x+8 \right)\left( {{x}^{3}}+7x+9 \right)={{x}^{5}}+7{{x}^{3}}+9{{x}^{2}}-5{{x}^{4}}-35{{x}^{2}}-45x+8{{x}^{3}}+56x+72$
$={{x}^{5}}-5{{x}^{4}}+15{{x}^{3}}-26{{x}^{2}}+9x+72$……………..(4)
Differentiating both the sides of eq. (4) with respect to x, we will get,
$\dfrac{d}{dx}\left( {{x}^{2}}-5x+8 \right)\left( {{x}^{3}}+7x+9 \right)=\dfrac{d}{dx}\left( {{x}^{5}}-5{{x}^{4}}+15{{x}^{3}}-26{{x}^{2}}+9x+72 \right)$
$=\dfrac{d}{dx}\left( {{x}^{5}} \right)-\dfrac{d}{dx}\left( 5{{x}^{4}} \right)+\dfrac{d}{dx}\left( 15{{x}^{3}} \right)-\dfrac{d}{dx}\left( 26{{x}^{2}} \right)+\dfrac{d}{dx}\left( 9x \right)+\dfrac{d}{dx}\left( 72 \right)$
$=5{{x}^{4}}-20{{x}^{3}}+45{{x}^{2}}-52x+9$………………….. (5)

(iii) Differentiation By logarithmic differentiation
Consider $h(x)=\left( {{x}^{2}}-5x+8 \right)\left( {{x}^{3}}+7x+9 \right)$……………… (6)
Now taking take logarithm with base e on both the sides of eq. (6), we will get,
{{\log }_{e}}h(x)={{\log }_{e}}\left\\{ \left( {{x}^{2}}-5x+8 \right)\left( {{x}^{3}}+7x+9 \right) \right\\}
$={{\log }_{e}}\left( {{x}^{2}}-5x+8 \right)+{{\log }_{e}}\left( {{x}^{3}}+7x+9 \right)$…………………………………… (7)
Differentiating both the sides of eq. (8) with respect to x, we will get,
\Rightarrow \dfrac{d}{dx}\left\\{ {{\log }_{e}}h(x) \right\\}=\dfrac{d}{dx}\left\\{ {{\log }_{e}}\left( {{x}^{2}}-5x+8 \right)+{{\log }_{e}}\left( {{x}^{3}}+7x+9 \right) \right\\}
\Rightarrow \dfrac{1}{h(x)}\dfrac{d}{dx}\left\\{ h(x) \right\\}=\dfrac{1}{\left( {{x}^{2}}-5x+8 \right)}\dfrac{d}{dx}\left( {{x}^{2}}-5x+8 \right)+\dfrac{1}{\left( {{x}^{3}}+7x+9 \right)}\dfrac{d}{dx}\left( {{x}^{3}}+7x+9 \right)
\Rightarrow \dfrac{d}{dx}\left\\{ h(x) \right\\}=\dfrac{h(x)}{\left( {{x}^{2}}-5x+8 \right)}\left( 2x-5 \right)+\dfrac{h(x)}{\left( {{x}^{3}}+7x+9 \right)}\left( 3{{x}^{2}}+7 \right)…………………………….. (8)
Substituting the value of h(x) from eq. (6) in eq. (8), we will get,
\Rightarrow \dfrac{d}{dx}\left\\{ \left( {{x}^{2}}-5x+8 \right)\left( {{x}^{3}}+7x+9 \right) \right\\}=\left( {{x}^{3}}+7x+9 \right)\left( 2x-5 \right)+\left( {{x}^{2}}-5x+8 \right)\left( 3{{x}^{2}}+7 \right)
$=2{{x}^{4}}-5{{x}^{3}}+14{{x}^{2}}-35x+18x-45+3{{x}^{4}}+7{{x}^{2}}-15{{x}^{3}}-35x+24{{x}^{2}}+56$
On solving, we get
\Rightarrow \dfrac{d}{dx}\left\\{ \left( {{x}^{2}}-5x+8 \right)\left( {{x}^{3}}+7x+9 \right) \right\\}=5{{x}^{4}}-20{{x}^{3}}+45{{x}^{2}}-52x+9
Here differentiating in the three ways mentioned we get the same answer.

Note: Differentiation of an exponent is given by$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$. Differentiation of a logarithmic function is given by$\dfrac{d}{dx}\left( {{\log }_{e}}u \right)=\dfrac{1}{u}\dfrac{du}{dx}$. Try not to make any calculation mistakes while solving the question and while doing differentiation, do differentiation carefully and systematically.