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How to find the value of $$f_x \ \text{and} \ f_y$$

If: $$f(x,y) = \int_y^x e^{t^2} \, dt$$

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  • $\begingroup$ (First) fundamental theorem of calculus. $\endgroup$ – peek-a-boo Jan 22 at 10:31
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The derivative of $\int_0^xf(t)dt$ is $f(x)$.

In this case, $f(x,y)=\int_0^x{t^2}dt-\int_0^ye^{t^2}dt$ We deduce that $f_x=e^{x^2}$ and $f_y=-e^{y^2}$.

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Let $y$ be constant , then we get, by the FTC:

$$f_x(x,y)= e^{x^2}.$$

Let $x$ be constant , then $f(x,y)=- \int_x^y e^{t^2}$ and we get, by the FTC:

$$f_y(x,y)= -e^{y^2}.$$

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Have you studied the fundamental theorem of calculus? It shows how to differentiate indefinite integrals. In this case,

$$ f_x = e^{x^2}, \qquad f_y = -e^{y^2}. $$

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    $\begingroup$ you're missing a minus sign for $f_y$. $\endgroup$ – peek-a-boo Jan 22 at 10:33
  • $\begingroup$ @peek-a-boo you are right, I'll correct. $\endgroup$ – PierreCarre Jan 22 at 11:31

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