Question: Find the partial derivatives, $f_x(x, y)$ and $f_y(x,y)$, of the function $$f(x,y)=\int_y^xcos(3t^2+9t-1)dt$$

My attempt is as follows.

  1. Substitution:




  2. Plug in $u$:



    and I don't know how to continue.

I've been stuck on this question for a few days now. Tried searching across all platforms but none has similar questions like this.

WolframAlpha shows an answer that involves the Fresnel C and S integrals but my class hasn't mentioned this anywhere.

  • $\begingroup$ Just use the Leibniz integral rule. You should get $$f_{x} = \cos(3x^{2}+9x-1)$$ and $$f_{y} = -\cos(3y^{2}+9y-1)$$ $\endgroup$ – mattos Apr 13 '19 at 4:54

No need to calculate the integral: $f_(x,y)=\cos(3x^{2}+9x-1)$ and $f_y((x,y)=-\cos(3y^{2}+9y-1)$. You only have to know that the derivative of an indefinite integral gives back the original function.

  • $\begingroup$ Oh thank you so much! So is it a general solution that $f_y (x,y)$, the lower limit, is just the original function with a negative sign? $\endgroup$ – Vero Apr 13 '19 at 8:01

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