If f(x) = cosx + sinx then dy/dx at x= 2pie/3 is
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f(x) = sinx + cosx
differentiate with respect to x
df(x)/dx = cosx - sinx
dy/dx at x = 2π/3
dy/dx = cos2π/3 - sin2π/3
= cos120° - sin120°
= -1/2 - √3/2
= - (1 + √3)/2
differentiate with respect to x
df(x)/dx = cosx - sinx
dy/dx at x = 2π/3
dy/dx = cos2π/3 - sin2π/3
= cos120° - sin120°
= -1/2 - √3/2
= - (1 + √3)/2
Answered by
0
Hey there !!!!!!
~~~~~~~~~~~~~~~~~~~~~
f(x) = cosx+sinx
Differentiating wrt to "x"
dy/dx = -sinx+cosx
dy/dx = cosx-sinx
At x= 2π/3
= cos2π/3-sin2π/3
= cos120-sin120
= -1/2-√3/2
=-(1+√3)/2 = -(1+1.7)/2 =-2.7/2 = -1.35
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hope this helped you..............
~~~~~~~~~~~~~~~~~~~~~
f(x) = cosx+sinx
Differentiating wrt to "x"
dy/dx = -sinx+cosx
dy/dx = cosx-sinx
At x= 2π/3
= cos2π/3-sin2π/3
= cos120-sin120
= -1/2-√3/2
=-(1+√3)/2 = -(1+1.7)/2 =-2.7/2 = -1.35
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hope this helped you..............
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