Find partial derivatives of function f(x,y) where f(x,y) is defined as an integral
Hint For simplicity, assume there is an anti-derivative $H(t)$ for $h(t)$. By the Fundamental Theorem of Calculus, $$ f(x,y) = \int_{y \sin (x^3)}^{x^3 \cos(y^2)} h(t) dt = H\left(x^3 \cos\left(y^2\right)\right) - H\left(y \sin \left(x^3\right)\right) $$ and now, use the chain rule: $$ \frac{\partial f}{\partial x} = H'\left(x^3 \cos\left(y^2\right)\right) \frac{\partial x^3 \cos\left(y^2\right)}{\partial x} - H'\left(y \sin \left(x^3\right)\right) \frac{\partial y \sin \left(x^3\right)}{\partial x} $$ and now use $H'=h$ and do the arithmetic...
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HDM
Computer Science and Engineering major with a minor in Mathematics.
Updated on August 01, 2022Comments
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HDM over 1 year
<. Find the partial derivatives <(x,y)< and <(x,y)<, in terms of the function <, where the function < is defined by the following integral.
<$$ f(x,y)=\int_a^bh(t) dt $$
$<$ a=ysin(x^3) $$ and $$ b=x^3cos(y^2) $$
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MPW over 9 yearsThe limits of integration are variable, so the entire expression is a function of $x$ and $y$. Thus it makes sense to find the indicated partials.
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MPW over 9 yearsThink of this as $f(x,y)=\int_{a(x,y)}^{b(x,y)}h(t)dt$.
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