# Find the derivative of the given function. [closed]

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

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

• (First) fundamental theorem of calculus. – peek-a-boo Jan 22 at 10:31

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 $$f_y=-e^{y^2}$$.

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:

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

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

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

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