For the given functions, (a) express startfraction dw over dt endfraction as a function of t, both by using the chain rule and by expressing w in terms of t and differentiating directly with respect to t. then (b) evaluate startfraction dw over dt endfraction at the given value of t. wequals6 y e superscript x baseline minus ln z, xequalsln (t squared plus 1 ), yequalstangent superscript negative 1 baseline t, zequalse superscript t; tequals1

se explanation, and the final value is 16.995574

Explanation:

a) As is a function of . In order to use chain rule first we compute the partial derivatives of   with respect to .

then the total derivatives of each term with respect to t:

we get:

next we replace x,y and  z with the respective functions:

we cancel out the terms at the end, e^ln(expression) always will be equal to expression

after some algebra we get:

b) as a function of t we directly replace  . and derivate with respect to t, before derivating, we can cancel out some terms:

now we replace t = 1

answer:

i lost in getting someone back.

i learned that people are really two-faced and no matter how angelic someone appears they always have a devil side and that you cant trust anyone anymore no matter how dear they are to you.

if your wondering who it waz it waz mah still miss her tho

~batmans wife dun dun