Question

find the derivative function of y = x✓x​

Answer 1

Solution :

Method (i) :

Given the function of y is $\sf{x\sqrt{x}}$ can also be written as,

$:\implies \sf{x\sqrt{x} = x^{1} \times x^{\tiny{1/2}}} \\ \\ :\implies \sf{x\sqrt{x} = x^{1 + 1/2}} \:\:\:\: \sf{[\because a^{n} \times a^{m} = a^{(n + m)}]} \\ \\ :\implies \sf{x\sqrt{x} = x^{(2 + 1)/2}} \\ \\ :\implies \sf{x\sqrt{x} = x^{\tiny{3/2}}} \\ \\ \boxed{\therefore \sf{x\sqrt{x} = x^{\tiny{3/2}}}}$

Hence the function of y is $\sf{x^{3/2}}$.

Now by differentiating the function with respect to x, we get :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d(x^{3/2})}{dx}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{3}{2} \times (x^{3/2 - 1})} \:\:\:\: \sf{[\because \dfrac{d(x^{n})}{dx} = n \times x^{(n - 1)}]} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{3}{2} \times (x^{(3 - 2)/2})} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{3}{2} \times (x^{1/2})} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{3}{2}x^{1/2}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{3}{2}\sqrt{x}} \\ \\ \boxed{\therefore \sf{\dfrac{dy}{dx} = \dfrac{3}{2}\sqrt{x}}}$

Thus the derivative of $\sf{x\sqrt{x}}$ is $\sf{\dfrac{3}{2}\sqrt{x}}$

Method (ii) :

Product rule of differentiation :

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

By using the product rule of differentiation and substituting the values in it, we get : [Here, u = x and v = √x]

$:\implies \sf{\dfrac{d(vu)}{dx} = u\dfrac{d(v)}{dx} + v\dfrac{d(u)}{dx}} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = (x)\dfrac{d(\sqrt{x})}{dx} + (\sqrt{x})\dfrac{d(x)}{dx}} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = (x)\dfrac{d(x^{1/2})}{dx} + (\sqrt{x})\dfrac{d(x)}{dx}} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = (x) \times \dfrac{1}{2} \times x^{(1/2 - 1)} + (\sqrt{x})(1)} \:\:\:\: \sf{[\because \dfrac{d(x^{n})}{dx} = n \times x^{(n - 1)}]} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = (x) \times \dfrac{1}{2} \times x^{(1/2 - 1)} + (\sqrt{x})(1)} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = (x) \times \dfrac{1}{2} \times x^{(1 - 2)/2} + \sqrt{x}} \\ \\:\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = (x^{1}) \times \dfrac{1}{2} \times x^{(-1)/2} + \sqrt{x}} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = \dfrac{1}{2} \times x^{(-1)/2 + 1} + \sqrt{x}} \:\:\:\: \sf{[\because a^{n} \times a^{m} = a^{(n + m)}]} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = \dfrac{1}{2} \times x^{(-1 + 2)/2} + \sqrt{x}} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = \dfrac{1}{2} \times x^{1/2} + \sqrt{x}} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = \dfrac{\sqrt{x}}{2} + \sqrt{x}} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = \dfrac{\sqrt{x} + 2\sqrt{x}}{2}} \\ \\ :\implies \sf{\dfrac{d(x\sqrt{x})}{dx} = \dfrac{3\sqrt{x}}{2}} \\ \\ \boxed{\therefore \sf{\dfrac{dy}{dx} = \dfrac{3}{2}\sqrt{x}}}$