Question

Differentiation of (2x+3) (8x^2+2x+7)

Class XI ​

Answer 1

To Find :

The derivative of $\bf{(2x + 3)(8x^{2} + 2x + 7)}$

Solution :

To find the differentiation , first let us find a single Equation to Derive.

Using the Equation and by finding the product we get :

$:\implies \bf{(2x + 3)(8x^{2} + 2x + 7)}$

$:\implies \bf{2x(8x^{2} + 2x + 7) + 3(8x^{2} + 2x + 7)}$

$:\implies \bf{16x^{3} + 4x^{2} + 14x + 24x^{2} + 6x + 21}$

$:\implies \bf{16x^{3} + (4x^{2} + 24x^{2}) + (14x + 6x) + 21}$

$:\implies \bf{16x^{3} + 28x^{2} + 20x + 21}$

$\therefore \bf{16x^{3} + 28x^{2} + 20x + 21}$

Hence, the Equation to be derived is $\bf{16x^{3} + 28x^{2} + 20x + 21}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀Differentiation :

Now , Using the general rule of differentiation , we get :

Here,

$\bf{f(x) = 16x^{3} + 28x^{2} + 20x + 21}$.

Now Substituting this in the $\bf{\dfrac{dy}{dx}}$ , we get :

$:\implies \bf{\dfrac{dy}{dx} = \dfrac{d(16x^{3} + 28x^{2} + 20x + 21)}{dx}}$

Now , by splitting it , we get :

$:\implies \bf{\dfrac{dy}{dx} = \dfrac{d(16x^{3})}{dx} + \dfrac{d(28x^{2})}{dx} + \dfrac{d(20x)}{dx} + \dfrac{d(21)}{dx}}$

$:\implies \bf{\dfrac{dy}{dx} = \dfrac{d(16x^{3})}{dx} + \dfrac{d(28x^{2})}{dx} + \dfrac{d(20x)}{dx} + 0}\:\:\:\bigg[\because \dfrac{d(c)}{dx} = 0)\bigg]$

$:\implies \bf{\dfrac{dy}{dx} = (3 \times 16 \times x^{2}) + (2 \times 28 \times x) + 20 + 0}$

$:\implies \bf{\dfrac{dy}{dx} = 48x^{2} + 56x + 20 + 0}$

$:\implies \bf{\dfrac{dy}{dx} = 48x^{2} + 56x + 20}$

$\therefore \boxed{\bf{\dfrac{dy}{dx} = 48x^{2} + 56x + 20}}$

Hence, the differentiation of

$\bf{(2x + 3)(8x^{2} + 2x + 7)}$ is

$\bf{\dfrac{dy}{dx} = 48x^{2} + 56x + 20}$