Question

find dy/dx where

y = (root x + 3)(x² + 6)​

Answer 1

Answer :

The Derivative of $\sf{y = (\sqrt{x} + 3)(x^{2} + 6)}$ is $\sf{2x^{\tiny{\dfrac{3}{2}}} + 6x + \dfrac{(x^{2} + 6)}{2\sqrt{x}}}$

Explanation :

To find :

The Derivative of :

• $\sf{y = (\sqrt{x} + 3)(x^{2} + 6)}$

Knowledge required :

• Product rule of differentiation :

$\boxed{\sf{\dfrac{d(vu)}{dx} = u\dfrac{d(v)}{dx} + v\dfrac{d(u)}{dx}}}$

• Exponent rule of differentiation :

$\boxed{\sf{\dfrac{d(x^{n})}{dx} = nx^{(n - 1)}}}$

• Differentiation of a constant term is 0.

Solution :

By using the the product rule of differentiation, we get :

$:\implies \sf{\dfrac{d(vu)}{dx} = u\dfrac{d(v)}{dx} + v\dfrac{d(u)}{dx}}$

Here,

• u = √x + 3
• v = x² + 6

By substituting the values of u and v in the equation, we get :

$:\implies \sf{\dfrac{d(vu)}{dx} = (\sqrt{x} + 3) \times \dfrac{d(x^{2} + 6)}{dx} + (x^{2} + 6) \times \dfrac{d(\sqrt{x} + 3)}{dx}}$

Now by differentiating u and v, we get :

Differentiation of u :

$:\implies \sf{\dfrac{d(u)}{dx} = \dfrac{d(\sqrt{x} + 3)}{dx}}$

$:\implies \sf{\dfrac{d(u)}{dx} = \dfrac{d(x^{\dfrac{1}{2}} + 3)}{dx}}$

$:\implies \sf{\dfrac{d(u)}{dx} = \dfrac{d(x^{\tiny{\dfrac{1}{2}}})}{dx} + \dfrac{d(3)}{dx}}$

$:\implies \sf{\dfrac{d(u)}{dx} = \dfrac{d(x^{\tiny{\dfrac{1}{2}}})}{dx} + 0}$

$:\implies \sf{\dfrac{d(u)}{dx} = \dfrac{1}{2} \times x^{\tiny{\dfrac{1}{2} - 1}}})$

$:\implies \sf{\dfrac{d(u)}{dx} = \dfrac{1}{2} \times x^{\tiny{\big(\dfrac{1 - 2}{2}}\big)}}$

$:\implies \sf{\dfrac{d(u)}{dx} = \dfrac{1}{2} \times x^{\tiny{\dfrac{-1}{2}}}}$

$:\implies \sf{\dfrac{d(u)}{dx} = \dfrac{1}{2} \times \dfrac{1}{x^{\tiny{\dfrac{1}{2}}}}}$

$:\implies \sf{\dfrac{d(u)}{dx} = \dfrac{1}{2} \times \dfrac{1}{\sqrt{x}}}$

$:\implies \sf{\dfrac{d(u)}{dx} = \dfrac{1}{2\sqrt{x}}}$

$\boxed{\therefore \sf{\dfrac{d(\sqrt{x} + 3)}{dx} = \dfrac{1}{2\sqrt{x}}}}$

Differentiation of v :

$:\implies \sf{\dfrac{d(v)}{dx} = \dfrac{d(x^{2} + 6)}{dx}}$

$:\implies \sf{\dfrac{d(v)}{dx} = \dfrac{d(x^{2})}{dx} + \dfrac{d(6)}{dx}}$

$:\implies \sf{\dfrac{d(v)}{dx} = \dfrac{d(x^{2})}{dx} + 0}$

$:\implies \sf{\dfrac{d(v)}{dx} = 2 \times x^{2 - 1}}$

$:\implies \sf{\dfrac{d(v)}{dx} = 2 \times x^{1}}$

$:\implies \sf{\dfrac{d(v)}{dx} = 2x}$

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

Now by substituting the Derivative of u and v in the equation, we get :

$:\implies \sf{\dfrac{d(vu)}{dx} = (\sqrt{x} + 3) \times \dfrac{d(x^{2} + 6)}{dx} + (x^{2} + 6) \times \dfrac{d(\sqrt{x} + 3)}{dx}}$

$:\implies \sf{\dfrac{d(vu)}{dx} = (\sqrt{x})2x + 2x(3) + \dfrac{(x^{2} + 6)}{2\sqrt{x}}}$

$:\implies \sf{\dfrac{d(vu)}{dx} = (x^{\tiny{\dfrac{1}{2}}})2x + 2x(3) + \dfrac{(x^{2} + 6)}{2\sqrt{x}}}$

$:\implies \sf{\dfrac{d(vu)}{dx} = 2x^{\big(\tiny{\dfrac{1}{2} + 1}\big)} + 6x + \dfrac{(x^{2} + 6)}{2\sqrt{x}}}$

$:\implies \sf{\dfrac{d(vu)}{dx} = 2x^{\big(\tiny{\dfrac{1 + 2}{2}}\big)} + 6x + \dfrac{(x^{2} + 6)}{2\sqrt{x}}}$

$:\implies \sf{\dfrac{d(vu)}{dx} = 2x^{\tiny{\dfrac{3}{2}}} + 6x + \dfrac{(x^{2} + 6)}{2\sqrt{x}}}$

$\boxed{\therefore \sf{\dfrac{d[(\sqrt{x} + 3)(x^{2} + 6)]}{dx} = 2x^{\tiny{\dfrac{3}{2}}} + 6x + \dfrac{(x^{2} + 6)}{2\sqrt{x}}}}$