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

eozoischgc 2021-12-07 Answered

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

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Expert Answer

ramirezhereva
Answered 2021-12-08 Author has 1526 answers

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,
\(\displaystyle{{f}_{{{x}}}{\left({x},{y}\right)}}={G}'{\left({x}\right)}={g{{\left({x}\right)}}}\)
\(\displaystyle{{f}_{{{y}}}{\left({x},{y}\right)}}={G}'{\left({y}\right)}=-{g{{\left({y}\right)}}}\)

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Kindlein6h
Answered 2021-12-09 Author has 3897 answers

\( \frac{d}{dx}\int_{\alpha (x)}^{\beta(x)} f(x,t)dt=f(x,\beta (x))-f(x,\alpha (x))+\int_{\alpha (x)}^{\beta(x)}\frac{\partial f(x,t)}{\partial x}dt\)
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).

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