Math, asked by srj100, 8 months ago

find dy/dx where

y = (root x + 3)(x² + 6)​

Answers

Answered by Anonymous
12

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}}}} \\ \\

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