Answer:
[tex](a) dy= 7sec^2(7t)dt\\\\(b) dy = \frac{-16v}{(4+v^2)^2} dv[/tex]
Step-by-step explanation:
Given;
(a) y = tan (7t)
let u = 7t
y = tan(u)
du/dt = 7
dy/du = sec²(u)
[tex]\frac{dy}{dt} = \frac{dy}{du} *\frac{du}{dt}\\\\\frac{dy}{dt} = sec^2(u)* 7\\\\\frac{dy}{dt} =7sec^2(u)\\\\\frac{dy}{dt} = 7sec^2(7t)\\\\dy = 7sec^2(7t)dt[/tex]
(b)
[tex]y = \frac{4-v^2}{4+v^2}\\\\[/tex]
let u = 4 - v²
du/dv = -2v
let v = 4 + v²
dv/du = 2v
[tex]y = \frac{4-v^2}{4+v^2}\\\\\frac{dy}{dv} = \frac{vdu-udv}{v^2} \\\\\frac{dy}{dv} =\frac{-2v(4+v^2)-2v(4-v^2)}{(4+v^2)^2}\\\\\frac{dy}{dv} =\frac{-8v-2v^3-8v+2v^3}{(4+v^2)^2}\\\\\frac{dy}{dv} =\frac{-16v}{(4+v^2)^2}\\\\dy = \frac{-16v}{(4+v^2)^2}dv[/tex]
