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.
1:f∘r(x)=ln(et⋅t)2:gradf(x,y)=[fx′fy′]=[1x1y]r′(t)=[(et)′t′]=[et1]3.1:(f∘r)′(t)=(t+lnt)′=1+1t3.2:(f∘r)′(t)=∂f∂x(r1(t),r2(t))∂r1∂t+∂f∂y(r1(t),r2(t))∂r2∂t==1r1(t)et+1r2(t)⋅1=etet+1t=1+1tAnswers are same1:\\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\,\,same1:f∘r(x)=ln(et⋅t)2:gradf(x,y)=[fx′fy′]=[x1y1]r′(t)=[(et)′t′]=[et1]3.1:(f∘r)′(t)=(t+lnt)′=1+t13.2:(f∘r)′(t)=∂x∂f(r1(t),r2(t))∂t∂r1+∂y∂f(r1(t),r2(t))∂t∂r2==r1(t)1et+r2(t)1⋅1=etet+t1=1+t1Answersaresame
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