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dudeperuvian

  • 2 years ago

if y=(√3x^2-4), then dy = a. 6x(√3x^2-4)dx b. (3x)/(√3x^2-4)dx c. (6x)/(√3x^2-4) d. 2(√3x^2-4)/(3x)dx there is an E choice but it cut off, is it E??...

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  1. vf321
    • 2 years ago
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    Dude, we're not here to do problems for you. I'm here to help work through them. Your question should be: Help me find the differential dy given \[y=\sqrt x^2-4\]

  2. vf321
    • 2 years ago
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    So do you need help differentiating or not? Also, your function's not clear. Is this it? \[y=\sqrt{3x^2-4}\]

  3. dudeperuvian
    • 2 years ago
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    yes vf321, sorry but \[y=\sqrt{3x^2-4}\] is right

  4. vf321
    • 2 years ago
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    When dealing with chain rule, it often helps to make your own functions up. Let: \[f(x) = \sqrt x\]\[g(x)=3x^2-4\]Then,\[y=f(g(x))\]Well let's take d/dx of both sides.\[\frac{dy}{dx}=f'(g(x))g'(x)\]Now, all you have to do is multiply by dx on both sides and replace all the f's and g's with their actual values. \[\frac{dy}{dx}dx=f'(g(x))g'(x)dx\]\[dy = f'(g(x))g'(x)dx\]

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