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lgbasallote

  • 3 years ago

LGBADERIVATIVE: \[\huge \frac{\text d}{\text dx} x^{e^{x^2}}\]

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  1. anonymous
    • 3 years ago
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    rewrite as \[e^{e^{x^2}\ln(x)}\] and use the chain rule

  2. lgbasallote
    • 3 years ago
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    do you mean \[\ln y = e^{x^2} \ln x?\]

  3. cinar
    • 3 years ago
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    \[\huge x^{e^{x^2}}*lnx* e^{x^2}*2x\]

  4. lgbasallote
    • 3 years ago
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    uhh how?

  5. cinar
    • 3 years ago
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    \[\huge y=a^{f(x)}\] \[\huge y'=a^{f(x)}*f'(x)*lna\]

  6. lgbasallote
    • 3 years ago
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    oh i see

  7. lgbasallote
    • 3 years ago
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    yeah that makes sense thanks

  8. cinar
    • 3 years ago
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    I made a mistake..

  9. mukushla
    • 3 years ago
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    x is not constant that formula is for constants

  10. cinar
    • 3 years ago
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    \[\huge x^{e^{x^2}}*\left( e^{x^2}*2x*lnx+\frac{e^{x2}}{x}\right)\]

  11. cinar
    • 3 years ago
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    @mukushla you're right, but still we need that formula..

  12. cinar
    • 3 years ago
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    |dw:1341688029055:dw|

  13. mukushla
    • 3 years ago
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    right

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