Differentiating an integral

  • Thread starter PhMichael
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  • #1
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Homework Statement



[tex]\int_{2}^{y(x)}e^{t^2+1}dt + \int_{x^2+3x}^{0}\frac{e^z}{1+z}=0[/tex]

I need so find [tex]y'(0)[/tex].


The Attempt at a Solution



[tex]\frac{d}{dx}\int_{2}^{y(x)}e^{t^2+1}dt =y'(x) \cdot e^{y(x)^2+1} [/tex]

[tex]\frac{d}{dx}\int_{x^2+3x}^{0}\frac{e^z}{1+z}=-\frac{e^{x^2+3x}}{x^2+3x+1}\cdot (2x+3)[/tex]

adding them and substituting x=0 yields:

[tex]y'(0) \cdot e^{y(0)^2+1} -3=0[/tex]

but how can i find y'(0) from them equation?
 

Answers and Replies

  • #2
557
1

Homework Statement



[tex]\int_{2}^{y(x)}e^{t^2+1}dt + \int_{x^2+3x}^{0}\frac{e^z}{1+z}=0[/tex]

I need so find [tex]y'(0)[/tex].


The Attempt at a Solution



[tex]\frac{d}{dx}\int_{2}^{y(x)}e^{t^2+1}dt =y'(x) \cdot e^{y(x)^2+1} [/tex]

[tex]\frac{d}{dx}\int_{x^2+3x}^{0}\frac{e^z}{1+z}=-\frac{e^{x^2+3x}}{x^2+3x+1}\cdot (2x+3)[/tex]

adding them and substituting x=0 yields:

[tex]y'(0) \cdot e^{y(0)^2+1} -3=0[/tex]

but how can i find y'(0) from them equation?


[tex]y'(x)= \frac{d}{dx}\int_{2}^{y(x)}e^{t^2+1}dt + \frac{d}{dx}\int_{x^2+3x}^{0}\frac{e^z}{1+z}[/tex]

Go with solving....

[tex]y'(0) = y'(0) \cdot e^{y(0)^2+1} -3[/tex]

[tex]y'(0) =\frac{-3}{1-e^{y(0)^2+1}} [/tex]

That's it.
 
  • #3
134
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but I need to find a number, not an expression involving y(0). how can y(0) be found?
 
  • #4
557
1
You don't know y(x), so you can't know y'(0).
 

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