## 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 2 years ago

1. apoorvk Group Title

Leibnitz rullllllleeeeeeeeeeeeeeeeee!!!!!!!!

2. apoorvk Group Title

$\frac{\int\limits_{h(x)}^{g(x)}f(t).dt}{dx}= f(g(x)).g'(x) - f(h(x)).h'(x)$ Here, $$f(t) = e^{t^2 + xt}, g(x)=\sin x\text{ and }h(x)=\cos x$$

3. apoorvk Group Title

$f'(x) = e^{\sin^2 x + x\sin x}.\cos x + e^{\cos^2 x + x\cos x}.\sin x$ $SOoooooOOOOoooooo,~ f'(0) = e^0+ 0 = \boxed{1}$

4. apoorvk Group Title

(i.e., if I did not make a characteristically stupid error again -_-)

5. Ishaan94 Group Title

Read the question again apoorv. $$\large e^{t^2 + xt}$$

6. apoorvk Group Title

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 Group Title

No, you are differentiating with respect to $$x$$.

8. apoorvk Group Title

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.