Question

Find the differential of each function. (a) y = tan( 5t ) dy = Correct: Your answer is correct. (b) y = 5 − v2 5 + v2

Answer 1

Answer:

$(a)$  $dy = 5sec^2(5t) \ dt$

$(a) \ dy = \frac{-20v}{(5+v^2)^2} \ dt$

Step-by-step explanation:

Given;

(a) y = tan(5t)

$(b) \ y = \frac{5-v^2}{5+v^2}$

Solving for (a)

y = tan(5t)

let u = 5t          

⇒y = tan(u)

du/dt = 5

dy/du = sec²u

$\frac{dy}{dt} =\frac{dy}{du} *\frac{du}{dt} \\\\\frac{dy}{dt} = sec^2(u)*5\\\\\frac{dy}{dt} = 5sec^2(u)\\\\\frac{dy}{dt} = 5sec^2(5t)$

$dy = 5sec^2(5t) \ dt$

Solving for b;

let u = 5 - v²

du/dv = -2v

let v = 5+ v²

dv/du = 2v

$\frac{dy}{dv} = \frac{vdu - udv}{v^2} \\\\\frac{dy}{dv} = \frac{-2v(5+v^2) - 2v(5-v^2)}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-10v-2v^3-10v+2v^3}{(5+v^2)^2}$

$\frac{dy}{dv} = \frac{-10v-10v}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-20v}{(5+v^2)^2}\\\\dy = \frac{-20v}{(5+v^2)^2} dt$