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anonymous

  • 5 years ago

find dx/dy for y^2 = 5x+2

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  1. anonymous
    • 5 years ago
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    Have you heard of implicit differentiation?

  2. anonymous
    • 5 years ago
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    yes i was wondering what the answer was as my teacher provided no answers. just wanted to double check =]

  3. anonymous
    • 5 years ago
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    Well, anyway, you need to take the derivative with respect to x on both sides. The left-hand side, being a function of x (i.e. y = y(x) will be differentiated using the chain rule). \[y^2=5x+2 \rightarrow \frac{d}{dx}y^2=\frac{d}{dx}(5x+2) \rightarrow2yy'=5\]\[y'=\frac{5}{2y} \rightarrow y'=\frac{5}{2\sqrt{5x+2}}, -\frac{5}{2\sqrt{5x+2}}\]

  4. anonymous
    • 5 years ago
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    Same answer?

  5. anonymous
    • 5 years ago
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    what is the difference between dx/dy and dy/dx?

  6. anonymous
    • 5 years ago
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    dy/dx is the rate of change in y with respect to x, and dx/dy is the rate of change of x with respect to y. In the first instance, we're measuring the change in y as a function of x (i.e. *we* change x and the function moves because of it). In the second, we're measuring the change in x as a function of y (i.e. *we* change y and x is forced to move because of it).

  7. anonymous
    • 5 years ago
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    Does that make sense?

  8. anonymous
    • 5 years ago
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    yes that does thank you very much! so for the question you just helped me out with you had found dx/dy right ?

  9. anonymous
    • 5 years ago
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    or did you find dy/dx?

  10. anonymous
    • 5 years ago
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    No...what you're technically doing is called an 'operation' in mathematics. You can move from one statement to another in mathematics and perform operations at each step, the only catch is that if you perform an operation on one side of an equation, you need to perform the operation on the other side.

  11. anonymous
    • 5 years ago
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    So, we needed the derivative of this equation with respect to x. I said we needed to take the derivative with respect to x on both sides.

  12. anonymous
    • 5 years ago
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    When we did it to the right-hand side, it was easy, since\[\frac{d}{dx}(5x+2)=5\]...you're used to that.

  13. anonymous
    • 5 years ago
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    But when we did it to the left-hand side, it wasn't so obvious as to what we were doing. What we're doing in implicit differentiation is using the chain rule. So, on the left-hand side,

  14. anonymous
    • 5 years ago
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    \[\frac{d}{dx}y^2=\frac{dy^2}{dy}.\frac{dy}{dx}=2y \frac{dy}{dx}=2yy'\]

  15. anonymous
    • 5 years ago
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    Do you see how we've used the chain rule? Nothing dodgy happened.

  16. anonymous
    • 5 years ago
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    awwww i see. hahahas i'm really stupid at maths. Well THANK YOU SO SO MUCH for your time and patience in explaining! =D

  17. anonymous
    • 5 years ago
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    np ;) and I'm pretty sure you're not stupid!

  18. anonymous
    • 5 years ago
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    You should be able to do your other problem now.

  19. anonymous
    • 5 years ago
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    THANK YOU THANK YOU !

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