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En

  • one year ago

PLEASE HEEEEELLLP!!! pls view it from the attachment. :)

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  1. anonymous
    • one year ago
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    What do you need help with?

  2. En
    • one year ago
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    i dont get the last part.

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  3. hartnn
    • one year ago
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    x^4 is divided to both the terms of the numerator

  4. hartnn
    • one year ago
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    |dw:1441522345977:dw| |dw:1441522367269:dw| and all other terms remain as is

  5. zepdrix
    • one year ago
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    \[\large\rm \frac{\frac{2x^3}{\sqrt{1-x^4}}-2x \arcsin(x^2)}{x^4}=\frac{\frac{2x^3}{\sqrt{1-x^4}}}{x^4}-\frac{2x \arcsin(x^2)}{x^4}\] \[\large\rm =\frac{2x^3}{x^4\sqrt{1-x^4}}-\frac{2x \arcsin(x^2)}{x^4}\] Do you understand these two steps?

  6. En
    • one year ago
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    yes ..

  7. En
    • one year ago
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    OKAY ! I GET IT NOW. THANKS!

  8. hartnn
    • one year ago
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    |dw:1441522888429:dw|

  9. En
    • one year ago
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    So you just multiply x^4 with sqrt(1+x^4) ?

  10. hartnn
    • one year ago
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    you solve like terms \(\large\rm =\frac{2\color{blue}{x^3}}{\color{blue}{x^4}\sqrt{1-x^4}}-\frac{2\color{blue}{x} \arcsin(x^2)}{\color{blue}{x^4}}\) and no need to touch other terms...

  11. hartnn
    • one year ago
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    multiplying x^4 with sqrt(1-x^4) will not have benefit for simplification

  12. En
    • one year ago
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    this part though...

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  13. hartnn
    • one year ago
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    ok, for simplicity, consider \(\sqrt {1-x^4} =A\) |dw:1441524054343:dw|

  14. En
    • one year ago
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    THANKS. :D

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is replying to Can someone tell me what button the professor is hitting...

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