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Y=θ[SIN(INθ)+COS(INθ)],THEN FIND dy÷dθ.
Solution) Y=θ[SIN(INθ)+COS(INθ)]applying u.v rulethen dy÷dθ={[ SIN(INθ)+COS(INθ) ] dθ÷dθ }+ {θ[ d÷dθ{SIN(INθ)+COS(INθ) ] } => SIN(INθ)+COS(INθ) + θ{ (COS(INθ)÷ θ) - (SIN(INθ)÷θ) } θ is canceled and sin(ln θ ) is also canceled then u will get => 2COS(INθ)
shape
Differentiate following. f ( x ) = sin (3x 2 + x ) Solution It looks as the outside function is the sine & the inside function is 3x 2 +x. The derivative is then.
i have problems with math and my teacher said that i am still progressing in math
2 times n times n divided by n
8.5cm square = m square
A plane is flying at 200 mph with a heading of 45degrees and encounters a wind mph from the west. What is the velocity and heading?
how do I solve these problems?
compare: 643,251: 633,512: 633,893. The answer is 633,512.
buyer behaviour
Get the Delta H (Enthalpy) and Delta V (Volume) of the both components below and compare by ratio. You need to use clapeyron equation and also need to draw the graphs. S A LG
y=Θ[sin(lnΘ)+cos(lnΘ)] dy/dΘ=[sin(lnΘ)+cos(lnΘ)] + Θ[cos(lnΘ)-sin(lnΘ)]*1/Θ ---->(Use Multiplication rule) dy/dΘ=2cosΘ.
y=Θ[sin(lnΘ)+cos(lnΘ)]
dy/dΘ=[sin(lnΘ)+cos(lnΘ)] + Θ[cos(lnΘ)-sin(lnΘ)]*1/Θ ---->(Use Multiplication rule)
dy/dΘ=2cosΘ.
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