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Home » Differentiate in three ways mentioned below: (i) by using product rule. (ii) by expanding the product to obtain a single polynomial.
Differentiate \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right) in three ways mentioned below: (i) by using product rule. (ii) by expanding the product to obtain a single polynomial.
CBSE | Class 12 | Continuity and Differentiability | Exercise 5.5 | Maths | NCERT
Differentiate \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right) in three ways mentioned below: (i) by using product rule. (ii) by expanding the product to obtain a single polynomial.

Solution:

Let’s take y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)

(i) By using the product rule:

\begin{aligned} &\frac{d y}{d x}=\left(x^{2}-5 x+8\right) \frac{d}{d x}\left(x^{3}+7 x+9\right)+\left(x^{3}+7 x+9\right) \frac{d}{d x}\left(x^{2}-5 x+8\right) \\ &\frac{d y}{d x}=\left(x^{2}-5 x+8\right)\left(3 x^{2}+7\right)+\left(x^{3}+7 x+9\right)(2 x-5) \end{aligned}

\frac{d y}{d x}=3 x^{4}+7 x^{2}-15 x^{3}-35 x+24 x^{2}+56+2 x^{4}-5 x^{3}+14 x^{2}-35 x+18 x-45

\frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}+11

(ii) To obtain a single polynomial expanding the product

y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)

y=x^{5}+7 x^{3}+9 x^{2}-5 x^{4}-35 x^{2}-45 x+8 x^{3}+56 x+72

y=x^{5}-5 x^{4}+15 x^{3}-26 x^{2}+11 x+ 72

\frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11

i

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