Question

Find the partial derivatives of the function.

\(F(\alpha ,\beta ) = \int\limits_\alpha ^\beta {\sqrt {{t^3} + 1} dt} \)

Short answer

Finding the first partial derivatives of the function.

\(f(\alpha ,\beta ) = \int\limits_\alpha ^\beta {\sqrt {{t^3} + 1} dt} \)

Step-by-step solution

Finding domain of the function \(f(x,y)\):

Consider the following functions \(g(t) = t + \ln t\)and\(f(x,y) = \frac{{1 - xy}}{{1 + {x^2}{y^2}}}\) to find composition of functions \(h(x,y) = g(f(x,y))\).

\(h(x,y) = g(f(x,y))\)

\( = g\left( {\frac{{1 - xy}}{{1 + {x^2}{y^2}}}} \right)\)

\(h(x,y) = \frac{{1 - xy}}{{1 + {x^2}{y^2}}} + \ln \left( {\frac{{1 - xy}}{{1 + {x^2}{y^2}}}} \right)\)

The domain of function \(f(x,y) = \frac{{1 - xy}}{{1 + {x^2}{y^2}}}\)is the set of all points in \({R^2}\) except denominator is zero.

But \(1 + {x^2}{y^2} > 0\), \(\forall (x,y) \in {R^2}\).

The domain of function \(f(x,y)\) is \({R^2}\).

Find points:

The domain of function \(g\) is equal to domain of logarithmic function is set of all positive real numbers.

The domain of \(h(x,y) = g(f(x,y))\)is the set of all points where it is \(f(x,y) > 0\)

\(\frac{{1 - xy}}{{1 + {x^2}{y^2}}} > 0\)

\(1 - xy > 0\)

Since \(1 + {x^2}{y^2} > 0\), \(\forall (x,y) \in {R^2}\).

The domain of the function \(h(x,y)\)is \(D = \left\{ {(x,y)|xy < 1} \right\}\).

The set of al points at which \(h(x,y)\) is continuous is \(D = \left\{ {(x,y)|xy < 1} \right\}\).