dy/dØ = cos(π/√(Ø^2+5) * (-1/2)(Ø^2 + 5)^(-3/2) ) * (2Ø)
=(-πØ/(Ø^2 + 5)^(3/2) * cos(π/√(Ø^2+5)
so when Ø = 2
dy/dØ = ...
You do the button - pushing.
Make sure to set your calculator to RAD
find dy/d(theta)
theta=2 ,
if y=sin(pi/(square root(theta^2+5)))
Show steps thanks!!!
3 answers
y = sin(π/√(θ^2+5))
weird, but just apply the rules
y = sun(u)
y' = cos(u) u'
u = π/v
u' = -π/v^2 v'
v = √(θ^2+5)
v' = θ/√(θ^2+5)
put it all together and it spells MOTHER. Or, more mathematically,
πθ*cos(π/√(θ^2+5)) / (θ^2+5)^(3/2)
weird, but just apply the rules
y = sun(u)
y' = cos(u) u'
u = π/v
u' = -π/v^2 v'
v = √(θ^2+5)
v' = θ/√(θ^2+5)
put it all together and it spells MOTHER. Or, more mathematically,
πθ*cos(π/√(θ^2+5)) / (θ^2+5)^(3/2)
y = sin [ pi*(t^2+5)^-.5 ]
y' = pi [cos pi*(t^2+5)^-.5] * [-.5(2t)](t^2+5)^-1.5
if t = 2
y' = pi {cos pi/ 3} *[-2] [ 1/27 ]
= -(2pi/27) cos (pi/3)
but cos pi/3 = 1/2
so
-pi/27
CHECK MY ARITHMETIC !!!!
=
y' = pi [cos pi*(t^2+5)^-.5] * [-.5(2t)](t^2+5)^-1.5
if t = 2
y' = pi {cos pi/ 3} *[-2] [ 1/27 ]
= -(2pi/27) cos (pi/3)
but cos pi/3 = 1/2
so
-pi/27
CHECK MY ARITHMETIC !!!!
=
0 answers