1
$\begingroup$

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:

    $u=3t^2+9t-1$

    $\frac{du}{dt}=6t+9$

    $dt=\frac{1}{6t+9}du$

  2. Plug in $u$:

    $\int_y^xcos(3t^2+9t-1)dt$

    $=\int_y^x\frac{cos(u)}{6t+9}du$

    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.

$\endgroup$
1
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.