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I am stuck trying to take the derivative of the following function:

$$F(t) = \int_{-2t^2}^{t^{1/2}}f(xt^{-1},x)dx$$

I am aware the fundamental theorem of calculus is relevant, but I am not sure with how to deal with a generic function $f(xt^{-1},x)$ like this.

Any help would be greatly appreciated!

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  • $\begingroup$ Maybe let $(u, v) = (xt^{-1}, x)$ and use the multivariate chain rule? $\endgroup$ Jul 9, 2017 at 5:22
  • $\begingroup$ If I try that, this is where I get to: $a = xt^{-1}$, $b = x$, the chain rule then says that $$\frac{\partial{f}}{\partial{t}} = \frac{\partial{f}}{\partial{a}}\frac{\partial{a}}{\partial{t}} + \frac{\partial{f}}{\partial{b}}\frac{\partial{b}}{\partial{t}}$$. Since $b$ isn't a function of t, then $$\frac{\partial{f}}{\partial{t}} = \frac{\partial{f}}{\partial{a}}\frac{\partial{a}}{\partial{t}} = -\frac{\partial{f}}{\partial{a}}\frac{x}{t^{2}}$$. When I put this back into the integral, I get $$\int_{-2t^2}^{t^{1/2}} -\frac{\partial{f}}{\partial{a}}\frac{x}{t^{2}} dx$$. Now what? $\endgroup$
    – lcoplan23
    Jul 9, 2017 at 5:59

1 Answer 1

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The FTC generalizes to the Leibniz integral rule: $$ \frac{d}{dx} \int_{a(x)}^{b(x)} g(x,t)dt = b'(x)g(x,b(x)) -a'(x)g(x,a(x)) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}g(x,t)dt.$$ We have $$\frac{\partial}{\partial t} f(xt^{-1},x) = -xt^{-2}f_1(xt^{-1},x) $$ where $f_1$ is the partial derivative of $f$ with respect to its first component. So, plugging everything in, we get $$ F'(t)=\frac{1}{2}t^{-1/2} f(t^{-1/2},t^{1/2})+4tf(-2t,-2t^2)-\int_{-2t^2}^{t^{1/2}}x t^{-2} f_1(xt^{-1},x)dx$$

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