# PSKf(x,y)=\int_{y}^{x}g(t)dtaskNSKg(t) continuous for all t. Evaluate the derivatives of f.

$$f(x,y)=\int_{y}^{x}g(t)dt$$
g(t) continuous for all t. Evaluate the derivatives of f.

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ramirezhereva

Let a be a real. Thus,
$$\displaystyle{f{{\left({x},{y}\right)}}}={\int_{{{a}}}^{{{x}}}}{g{{\left({t}\right)}}}{\left.{d}{t}\right.}-{\int_{{{a}}}^{{{y}}}}{g{{\left({t}\right)}}}{\left.{d}{t}\right.}={G}{\left({x}\right)}-{G}{\left({y}\right)}$$
$$G(X)=\int_{a}^{X} g(t)dt$$
$$\displaystyle{G}'{\left({X}\right)}={g{{\left({X}\right)}}}$$
Since g is continuous at R,
$$\displaystyle{{f}_{{{x}}}{\left({x},{y}\right)}}={G}'{\left({x}\right)}={g{{\left({x}\right)}}}$$
$$\displaystyle{{f}_{{{y}}}{\left({x},{y}\right)}}={G}'{\left({y}\right)}=-{g{{\left({y}\right)}}}$$

###### Not exactly what youâ€™re looking for?
Kindlein6h

$$\frac{d}{dx}\int_{\alpha (x)}^{\beta(x)} f(x,t)dt=f(x,\beta (x))-f(x,\alpha (x))+\int_{\alpha (x)}^{\beta(x)}\frac{\partial f(x,t)}{\partial x}dt$$
f(x,t) is a function of t only, the upper bound on the integral is just x and the lower bound just y. So the derivative of the integral with respect to x is g(x) and the derivative with respect to y is −g(y).