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son4ous [18]
3 years ago
9

Differentiate

rmula1" title=" \sqrt{x} + \frac{1}{5x} - 3" alt=" \sqrt{x} + \frac{1}{5x} - 3" align="absmiddle" class="latex-formula">
with respect to their variable ​
Mathematics
1 answer:
Leviafan [203]3 years ago
3 0

\\ \tt\longmapsto \dfrac{d}{dx}\sqrt{x}+\dfrac{1}{5x}-3

\\ \tt\longmapsto \dfrac{d}{dx}x^{\frac{1}{2}}+5x^{-1}-3

  • Derivative of any constant is 0

\\ \tt\longmapsto \dfrac{d}{dx}x^{\frac{1}{2}}+5x^{-1}

\boxed{\sf \dfrac{d(x^n)}{dx}=nx^{n-1}}

\\ \tt\longmapsto \dfrac{1}{2}x^{\frac{-1}{2}}-5x^{-2}

\\ \tt\longmapsto \dfrac{x}{2}^{\frac{-1}{2}}-\dfrac{1}{5x^2}

\\ \tt\longmapsto -\sqrt{\dfrac{x}{2}}-\dfrac{1}{5x^2}

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Answer:

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Step-by-step explanation:

The complete question is attached.

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Let x be the number of slices that Paul can afford to buy.

Weight of on slice is 0.04 pounds. So, weight of x slices is 0.04x pound.

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So, total cost of cheese for x slices = $5.99 × 0.04x

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