Answer to Question #328181 in Calculus for Dudu

Question

Let F be the R²-R function defined by f(x,y)=Inxy and let r be the R-R² function defined by r(t)=(e^t;t).

1.determine the composite function F o r: (simplify your answer).

2.determine gradf (x,y) and r'(t).

3.determine the derivative function (f o r)' by

3.1.differentiating the expression obtained in (1).

3.2.using the chain rule (theorem ) compare your answer.

Answer 1

$1:\\f\circ r\left( x \right) =\ln \left( e^t\cdot t \right) \\2:\\gradf\left( x,y \right) =\left[ \begin{array}{c} f'_x\\ f'_y\\\end{array} \right] =\left[ \begin{array}{c} \frac{1}{x}\\ \frac{1}{y}\\\end{array} \right] \\r'\left( t \right) =\left[ \begin{array}{c} \left( e^t \right) '\\ t'\\\end{array} \right] =\left[ \begin{array}{c} e^t\\ 1\\\end{array} \right] \\3.1:\\\left( f\circ r \right) '\left( t \right) =\left( t+\ln t \right) '=1+\frac{1}{t}\\3.2:\\\left( f\circ r \right) '\left( t \right) =\frac{\partial f}{\partial x}\left( r_1\left( t \right) ,r_2\left( t \right) \right) \frac{\partial r_1}{\partial t}+\frac{\partial f}{\partial y}\left( r_1\left( t \right) ,r_2\left( t \right) \right) \frac{\partial r_2}{\partial t}=\\=\frac{1}{r_1\left( t \right)}e^t+\frac{1}{r_2\left( t \right)}\cdot 1=\frac{e^t}{e^t}+\frac{1}{t}=1+\frac{1}{t}\\Answers\,\,are\,\,same$