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

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apoorvkBest ResponseYou've already chosen the best response.3
Leibnitz rullllllleeeeeeeeeeeeeeeeee!!!!!!!!
 one year ago

apoorvkBest ResponseYou've already chosen the best response.3
\[\frac{\int\limits_{h(x)}^{g(x)}f(t).dt}{dx}= f(g(x)).g'(x)  f(h(x)).h'(x)\] Here, \(f(t) = e^{t^2 + xt}, g(x)=\sin x\text{ and }h(x)=\cos x\)
 one year ago

apoorvkBest ResponseYou've already chosen the best response.3
\[f'(x) = e^{\sin^2 x + x\sin x}.\cos x + e^{\cos^2 x + x\cos x}.\sin x \] \[SOoooooOOOOoooooo,~ f'(0) = e^0+ 0 = \boxed{1}\]
 one year ago

apoorvkBest ResponseYou've already chosen the best response.3
(i.e., if I did not make a characteristically stupid error again _)
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
Read the question again apoorv. $$\large e^{t^2 + xt}$$
 one year ago

apoorvkBest ResponseYou've already chosen the best response.3
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..
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
No, you are differentiating with respect to \(x\).
 one year ago

apoorvkBest ResponseYou've already chosen the best response.3
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.
 one year ago
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