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NickAlmighty22go
Answered

I need help finding the Derivative of [tex]f(x)=x^{2} -\frac{1}{2} \sqrt{x}[/tex] with solution please

Sagot :

Karenalbina26go

Answer:

x1=0 x2= ³√2/2

Step-by-step explanation:

[tex]f(x)=x^{2} -\frac{1}{2} \sqrt{x}[/tex]

[tex] {x}^{2} - \frac{1}{2} \times \sqrt{x} [/tex]

[tex]{x}^{2} - \frac{1}{2} \times \sqrt{x} = 0[/tex]

[tex]2 {x}^{2} - \sqrt{x \: }= 0[/tex]

[tex] - \sqrt{x} = - 2 {x}^{2} [/tex]

[tex] \sqrt{x } = 2 {x}^{2} [/tex]

[tex]x = 4 {x}^{4} [/tex]

[tex]x - 4 {x}^{4} = 0[/tex]

[tex]x(1 - 4 {x}^{3} ) = 0[/tex]

[tex]x = 0[/tex]

[tex]1 - 4 {x}^{3} = 0[/tex]

[tex]x = 0[/tex]

[tex]x = \sqrt[3]{2} \div 2[/tex]

Hope it helps have a nice day po☺️❤️

XmamonskieXgo

finding the Derivative:

[tex]f(x) = {x}^{2} - \frac{1}{2} \sqrt{x} \\⟹f(x) = \frac{d}{dx}( {x}^{2} - \frac{1}{2} \times \sqrt{x}) \\⟹f'(x) = \frac{d}{dx}( {x}^{2}) + \frac{d}{dx}( - \frac{1}{2} \times \sqrt{x} \\⟹f'(x) = 2x + \frac{d}{dx} ( - \frac{1}{2} \times \sqrt{x} \\ ⟹f'(x) = 2x - \frac{1}{2} \times \frac{1}{2 \sqrt{x}} \\⟹ \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \\⟹{\boxed{f'(x) = 2x - \frac{1}{4 \sqrt{x}}}}[/tex]

The Derivative is

[tex]{\boxed{f'(x) = 2x - \frac{1}{4 \sqrt{x}}}}[/tex]

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