Question

Differentiate the following w.r.t.x and find the value when x=3; (6x^2+3)(3x+7)

Answer 1

Solution :

Knowledge required :

• Exponential rule of differentiation :

If y = x^n then the Differentiation of x^n will be n·x^(n - 1)

• Differentiation of constant term is 0.i.e,

d(7)/dx = 0

• Exponent rule that , x⁰ = 1.

Now,

First let us find the Differentiation of :

(6x² + 3)(3x + 7)

We know the product rule of differentiation,i.e,

$\boxed{\rm{\dfrac{d(uv)}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}}}$

Now by using the product rule and substituting the values in it, we get :

Here,

• u = 6x² + 3
• v = 3x + 7

$:\implies \rm{\dfrac{d(uv)}{dx} = u\dfrac{d(v)}{dx} + v\dfrac{d(u)}{dx}}$

$:\implies \rm{\dfrac{dy}{dx} = (6x^{2} + 3) \times \dfrac{d(3x + 7)}{dx} + (3x + 7) \times \dfrac{d(6x^{2} + 3)}{dx}}$ ⠀⠀⠀⠀⠀⠀⠀⠀Eq.(i)

Now let's Differentiate the term u i.e, (6x² + 3)

$:\implies \bf{\dfrac{dy}{dx} = \dfrac{d(6x^{2} + 3)}{dx}}$

$:\implies \bf{\dfrac{dy}{dx} = 2 \times 6x^{(2 - 1)} + 0}$

$:\implies \bf{\dfrac{dy}{dx} = 2 \times 6x^{1}}$

$:\implies \bf{\dfrac{dy}{dx} = 12x}$

$\boxed{\therefore \bf{\dfrac{d(6x^{2} + 3)}{dx} = 12x}}$

Hence the Differentiation of 6x² + 3 is 12x.

Now let's Differentiate the term u i.e, (3x + 7)

$:\implies \bf{\dfrac{dy}{dx} = \dfrac{d(3x + 7)}{dx}}$

$:\implies \bf{\dfrac{dy}{dx} = 1 \times 3x^{(1 - 1)} + 0}$

$:\implies \bf{\dfrac{dy}{dx} = 1 \times 3x^{0}}$

$:\implies \bf{\dfrac{dy}{dx} = 3 \times 1}$

$:\implies \bf{\dfrac{dy}{dx} = 3}$

$\boxed{\therefore \bf{\dfrac{d(3x + 7)}{dx} = 3}}$

Hence the Differentiation of 3x + 7 is 3.

By substituting them in the equation (i) , we get :

$:\implies \rm{\dfrac{dy}{dx} = (6x^{2} + 3) \times \dfrac{d(3x + 7)}{dx} + (3x + 7) \times \dfrac{d(6x^{2} + 3)}{dx}}$

$:\implies \rm{\dfrac{dy}{dx} = (6x^{2} + 3) \times 3 + (3x + 7) \times 6x}$

$:\implies \rm{\dfrac{dy}{dx} = 18x^{2} + 9 + 18x^{2} + 42x}$

$:\implies \rm{\dfrac{dy}{dx} = 36x^{2} + 42x + 9}$

$\boxed{\therefore \rm{\dfrac{dy}{dx} = 36x^{2} + 42x + 9}}$

Hence the derivative of (6x² + 3)(3x + 7) is 36x² + 42x + 9.

Now by substituting the value of x in the equation , 36x² + 42x + 9 , we get :

==> 36 × (3)² + 42 × 3 + 9

==> 36 × 9 + 126 + 9

==> 323 + 126 + 9

==> 459

Hence the value of (6x² + 3)(3x + 7) (After Differentiating) is 459.