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Ishaan94

  • 3 years ago

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

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  1. apoorvk
    • 3 years ago
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    Leibnitz rullllllleeeeeeeeeeeeeeeeee!!!!!!!!

  2. apoorvk
    • 3 years ago
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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, \(f(t) = e^{t^2 + xt}, g(x)=\sin x\text{ and }h(x)=\cos x\)

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

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

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

  6. apoorvk
    • 3 years ago
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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..

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

  8. apoorvk
    • 3 years ago
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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.

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