Question

Let f : Rn × Rn → R be the inner product function: f(r,y)-(2,3) 1. Using the definition of multivariable derivative, calculate D fab and the Jacobian matrix f'(a, b) 2. If f, g : Rn → R are differentiable and h : R → R is defined by h(t)-(f(t), g(t)), show that 3. If f : R → Rn is differentiable and Ilf(t)ll = 1 for all t, show that(f,(t)T,f(t))-0

Answer 1

Let f:Rⁿ×Rⁿ--->R be the inner product function :f(x,y)=(x,y).

(1). Multivariable derivative is a combination of derivative of different variable functions.

First of all,In vector calculus ,Jacobian matrix is the matrix of all first order partial derivatives of a vector valued function.when the matrix is a square matrix,both the matrix and its determinant are referred to as the Jacobian.

Here,By definition of the multi variable derivative function Df_a,b are the different derivative functiond of the various elements used to form jacobian matrix f^'(a,b).

Df_a,b=(df_a,b/dx₁,df_a,b/dx₂,........df_a,b/dx_n).

And the Jacobian matrix formed with above multi variable derivatives is

J=[df/dx₁.......df/dx]=f^'(a,b)=[df₁(a,b)/dx₁..........df₁/dx_n]

.

.

[df_m(a,b)/dx₁.....df_m(a,b)/dx_n].

(2).If f,g:Rⁿ--->R are differentiable and h:R--->R is defined by h(t)=(f(t),g(t)) we have to find out the derivative of h(t) which is a combination of two different functions.So,we have to apply derivative formula of u,v that means

d(u,v)=(u',v)+(u,v^').

By applying above approximations

h^'(a)=D(f(a),g(a)).

h'(a)=(f^'(a),g(a))+(f(a),g^'(a)).

Here,in the Jacobian matrix form,square symmetric that means normal matrix equal to transpose of that matrix.

So,Finally from above explanations, It is proved that h'(a)=(f^'(a)^T,g(a))+(f(a),g^'(a)^T).

(3).If R--->Rⁿ is diffeentiable and ||f(t)|| =1 for all t,then derivative function of f(t) definitely equal to zero because derivative of constant is zero and here given that f(t) does not contain any variable and this value is 1.

And transpose of that function also equal to zero.

So,From above explanations (f^'(t)^T,f(t))=0.because the function with zero also equal to zero.