Question

Find the partial derivatives of the function.

\(F(x,y) = \int\limits_x^y {cos({e^t})dt} \)

Short answer

Finding the first partial derivatives of the function.

\(F(x,y) = \int\limits_x^y {\cos ({e^t})dt} \)

Step-by-step solution

Given data:

\(g(t) = {(t)^2} + \sqrt t \)

\(g(t) = {t^2} + \sqrt t \) and \(f(x) = 2x + 3y - 6\)

Finding \(h(x,y)\)

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

\( = (2x + 3y - 6) + \sqrt {2x + 3y - 6} \)

\( = (2x)2 + (3y)2 + ( - 6)2 + 2(2x)(3y) - 2(3y)(6) - (6)(2x) + \sqrt {2x + 3y - 6} \)

\( \Rightarrow h(x,y) = 4{x^2} + 9{y^2} + 36 + 12xy - 24x - 36y + \sqrt {2x + 3y - 6} \)

i.e,

Now \(h(x,y)\)is defined everywhere except for the values of \(x\) and \(y\):

for which \(2x + 3y - 6 < 0\)

Hence \(h(x,y)\) is continuous on the set.

\(\left\{ {(x,y):2x + 3y - 6 \ge 0} \right\}\)

Therefore,