How do you differentiate #f(x)=(x+1)/sqrtx# using the quotient rule?

Answer 1

#f'(x)=(x-1)/(xsqrtx)#

Let #f(x)=(x+1)/sqrtx#. To find #f'#, we will use the quotient rule
#d/dxu/v=(vu'-uv')/v^2#

So

#f'(x)=d/dx(x+1)/sqrtx=(sqrtxd/dx(x+1)-(x+1)d/dxsqrtx)/(sqrtx)^2=(sqrtx-(x+1)/(2sqrtx))/x=(2x-x-1)/(2xsqrtx)=(x-1)/(2xsqrtx)#
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Answer 2

To differentiate ( f(x) = \frac{x+1}{\sqrt{x}} ) using the quotient rule:

  1. Identify the numerator function: ( u(x) = x + 1 )
  2. Identify the denominator function: ( v(x) = \sqrt{x} )
  3. Apply the quotient rule formula: [ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} ]
  4. Calculate the derivatives: [ u'(x) = 1 ] [ v'(x) = \frac{1}{2\sqrt{x}} ]
  5. Plug the derivatives and functions into the quotient rule formula: [ f'(x) = \frac{(1)(\sqrt{x}) - (x+1)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2} ]
  6. Simplify the expression: [ f'(x) = \frac{\sqrt{x} - \frac{x+1}{2\sqrt{x}}}{x} ] [ f'(x) = \frac{2x\sqrt{x} - (x+1)}{2x\sqrt{x}} ] [ f'(x) = \frac{2x\sqrt{x} - x - 1}{2x\sqrt{x}} ]

So, the derivative of ( f(x) = \frac{x+1}{\sqrt{x}} ) using the quotient rule is ( f'(x) = \frac{2x\sqrt{x} - x - 1}{2x\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.

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.

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.

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.

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