Partial Derivatives with Integral

Question

$$f(x,y) = \int\limits_y^x \cos(7t^2 - 1t + 2) dt$$

How do I get these partial derivatives:

$$f_x (x,y)\ \text{ and } \ f_y (x,y) ?$$

Answer 1

Use the fundamental theorem of calculus: $$ \frac{\partial}{\partial x}\int_y^x \cos(7\,t^2 - t + 2)\,dt=\cos(7\,x^2 - x + 2). $$ Can you do the partial with respect to $y$?

Answer 2

Let:

$$F(t) = \int \cos(7t^2 - t + 2) dt$$

then:

$$f(x,y) = \int_y^x \cos(7t^2 - t + 2) dt = F(x) - F(y)$$

and so:

$$f_x (x,y) = F'(x) = \cos(7x^2 - x + 2)$$ $$f_y (x,y) = -F'(y) = -\cos(7y^2 - y + 2)$$

Answer 3

The integrand $\cos(7t^{2}-t+2)$ is smooth, one does without worry by the Fundamental Theorem of Calculus to get that $\dfrac{\partial}{\partial y}\displaystyle\int_{y}^{x}\cos(7t^{2}-t+2)dt=-\cos(7y^{2}-y+2)$.

Answer 4

Note that in general

$$f(t)=\int_{a(t)}^{b(t)}g(u) du\implies f'(t)=g(b(t))\cdot b'(t)-g(a(t))\cdot a'(t)$$

thus

$$f(x,y) = \int_y^x \cos(7t^2 - 1t + 2) dt\implies f_x=\cos(7x^2 - 1x + 2)$$

$$f(x,y) = \int_y^x \cos(7t^2 - 1t + 2) dt\implies f_y=-\cos(7y^2 - 1y + 2)$$

Answer 5

$$\int_y^x g(t)\,dt=G(x)-G(y),$$ where $G'(t)=g(t)$. So

$$\frac{\partial f(x,y)}{\partial x}=G'(x)-0=g(x),\\ \frac{\partial f(x,y)}{\partial y}=0-G'(y)=-g(y). $$