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Differentiation of Exponential Functions - Question watch

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    How do I go about differentiating xex^2 ?

    I started by letting u = x and v = ex^2 , so du/dx = 1 but I don't know how to find dv/dx.

    Any help would be greatly appreciated!

    Thanks!
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    (Original post by grace15)
    How do I go about differentiating xex^2 ?

    I started by letting u = x and v = ex^2 , so du/dx = 1 but I don't know how to find dv/dx.

    Any help would be greatly appreciated!

    Thanks!
    Use the chain rule on e^{x^2}

    Let w=x^2
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    (Original post by ghostwalker)
    Use the chain rule on e^x^2
    So if I let y= (ex )2 then dy/dx = 2 (ex )(ex ) = 2ex^2 ?
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    So, you have to do x' * (e^(x^2)) + x*((e^(x^2))'
    Its the formula u'v + uv'
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    (Original post by grace15)
    So if I let y= (ex )2 then dy/dx = 2 (ex )(ex ) = 2ex^2 ?
    I edited my previous post to suggest the substitution.


    NOTE: e^{x^2}\not=(e^x)^2

    Edit:

    (e^x)^2 =e^x\times e^x = e^{2x}
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    So if I let v=ew and let w = x2 then dy/dx of ex^2 is 2xex^2 ?

    Then would I use the product rule to get dy/dx = x(2xex^2 ) + ex^2 ?

    Which would simplify to ex^2 (2x2 + 1) ?

    Is this correct?
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    Yes.
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    (Original post by grace15)
    So if I let v=ew and let w = x2 then dy/dx of ex^2 is 2xex^2 ?
    Yes.

    Then would I use the product rule to get dy/dx = x(2xex^2 ) + ex^2 ?

    Which would simplify to ex^2 (2x2 + 1) ?

    Is this correct?
    And yes.
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    (Original post by ghostwalker)
    Yes.



    And yes.
    Great! Thank you very much for your help!
 
 
 
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