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How do you differentiate #y = -(1/x^2) - sqrtx + (1/2)#?

Answer 1

Aurora Alvarado

#y'=2/x^3-1/(2*sqrt(x))#

Writing #y=-x^(-2)-x^(1/2)+1/2#

After the rule #(x^n)'=nx^(n-1)#

we get

#y'=2x^(-3)-1/2x^(-1/2)#

#y'=2/x^3-1/(2sqrt(x))#

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Answer 2

Isaiah Allen

#dy/dx=2/x^3-1/(2sqrtx)#

#"differentiate using the "color(blue)"power rule"#

#•color(white)(x)d/dx(ax^n)=nax^(n-1)#

#y=-x^-2-x^(1/2)+1/2#

#dy/dx=2x^-3-1/2x^(-1/2)+0#

#color(white)(dy/dx)=2/x^3-1/(2sqrtx)#

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Answer 3

Samantha Adkison

To differentiate the function ( y = -\frac{1}{x^2} - \sqrt{x} + \frac{1}{2} ), you would take the derivative of each term separately using the rules of differentiation:

For the term ( -\frac{1}{x^2} ), use the power rule for differentiation: ( \frac{d}{dx}\left(-\frac{1}{x^2}\right) = 2x^{-3} ).
For the term ( -\sqrt{x} ), use the power rule for differentiation and the chain rule: ( \frac{d}{dx}\left(-\sqrt{x}\right) = -\frac{1}{2\sqrt{x}} ).
For the constant term ( \frac{1}{2} ), its derivative is zero since it has no ( x ) term.

Adding these derivatives together, you get the derivative of the function:

[ \frac{dy}{dx} = 2x^{-3} - \frac{1}{2\sqrt{x}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.