Question

Use the chain rule for differentiation of vector-valued functions to compute d dt F(g(t)) for the indicated function. (Your instructors prefer angle bracket notation < > for vectors.) Frightarrowhead.gif(u) = and u = ln(t4)

Answer 1

Answer:

dF(g(t)/dt =  < (4/t) , (16/t)*ln(t) , (192/t)*[ln(t)]² >

Step-by-step explanation:

if g(t)=u=ln(t⁴) = 4*ln(t) and F(u)= <u,u²/2,u³/3> , then

F(g(t))=< ln(t⁴),[ln(t⁴)]²/2,[ln(t⁴)]³/3>

dF(g(t)/dt = dF(u(t))/du * du(t)/dt = < 1 , [ln(t⁴)] , 3*[ln(t⁴)]² > * (4/t) =   < 1 * (4/t), [ln(t⁴)] * (4/t) , 3*[ln(t⁴)]² * (4/t) >  = < (4/t) , (16/t)*ln(t) , (192/t)*[ln(t)]² >

then

dF(g(t)/dt =  < (4/t) , (16/t)*ln(t) , (192/t)*[ln(t)]² >

Answer 2

In this exercise we have to use our knowledge of vectors and indicate the function that will be formed, so we have to:

$dF(g(t)/dt = < (4/t) , (16/t)*ln(t) , (192/t)*[ln(t)]^2 >$

Then given that the functions of g(t) and f(t) as:

$g(t)=u=ln(t^4) = 4*ln(t) \\ F(u)= $

Combining the two functions we find an equation like:

$F(g(t))=< ln(t^4),[ln(t^4)]^2/2,[ln(t^4)]^3/3>$

Now solving differently, we have that results in:

$dF(g(t)/dt = dF(u(t))/du * du(t)/dt \\= < 1 , [ln(t^4)] , 3*[ln(t^4)]^2 > * (4/t) \\= < 1 * (4/t), [ln(t^4)] * (4/t) , 3*[ln(t^4)]^2 * (4/t) >\\dF(g(t)/dt = < (4/t) , (16/t)*ln(t) , (192/t)*[ln(t)]^2 >$

See more about vectores at brainly.com/question/13188123