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Ishaan94 Group Title

Compute \(f'\left(0\right)\). $$\large f\left(x\right) = \int\limits_{\cos x}^{\sin x} e^{t^2+xt}dt.$$

  • 2 years ago
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  1. apoorvk Group Title
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    Leibnitz rullllllleeeeeeeeeeeeeeeeee!!!!!!!!

    • 2 years ago
  2. apoorvk Group Title
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    \[\frac{\int\limits_{h(x)}^{g(x)}f(t).dt}{dx}= f(g(x)).g'(x) - f(h(x)).h'(x)\] Here, \(f(t) = e^{t^2 + xt}, g(x)=\sin x\text{ and }h(x)=\cos x\)

    • 2 years ago
  3. apoorvk Group Title
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    \[f'(x) = e^{\sin^2 x + x\sin x}.\cos x + e^{\cos^2 x + x\cos x}.\sin x \] \[SOoooooOOOOoooooo,~ f'(0) = e^0+ 0 = \boxed{1}\]

    • 2 years ago
  4. apoorvk Group Title
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    (i.e., if I did not make a characteristically stupid error again -_-)

    • 2 years ago
  5. Ishaan94 Group Title
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    Read the question again apoorv. $$\large e^{t^2 + xt}$$

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  6. apoorvk Group Title
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    I am sorry, but isn't that 'x' supposed to be taken as a constant when you integrate wrt to 't'?? I can't figure out where am going wrong. Or may be I do..

    • 2 years ago
  7. Ishaan94 Group Title
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    No, you are differentiating with respect to \(x\).

    • 2 years ago
  8. apoorvk Group Title
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    Well that's the later part of the thing, but when am integrating, 'x' is a constant wrt 't', no balls with that. After am done the integration thingy, I am supposed to put in limits, and BAZINGA! the function is now in terms of 'x' completely. NOW we differentiate. We could follow all these steps, except that it would take ages for me to integrate that thing. So, Leibnitz to the rescue. Either this, or am missing your whole point.

    • 2 years ago
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