Differentiate sin inverse sinx+cosx / root 2.
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y = arcsin(sin(x) + cos(x)/sqrt(2))
sin(y) = sin(x) + cos(x)/sqrt(2)
cos(y) * dy = cos(x) * dx - sin(x) * dx / sqrt(2)
sqrt(1 - sin(y)^2) * dy = dx * (cos(x) - sin(x)/sqrt(2))
sqrt(1 - sin(x)^2 - 2sin(x)cos(x)/sqrt(2) - cos(x)^2 / 2) * dy = dx * (cos(x) - sin(x)/sqrt(2))
sqrt((1/2) * cos(x)^2 - sqrt(2) * sin(x)cos(x)) * dy = dx * (cos(x) - sin(x)/sqrt(2))
dy/dx = (cos(x) - sin(x)/sqrt(2)) / sqrt((1/2) * cos(x)^2 - sqrt(2) * sin(x)cos(x))
sin(y) = sin(x) + cos(x)/sqrt(2)
cos(y) * dy = cos(x) * dx - sin(x) * dx / sqrt(2)
sqrt(1 - sin(y)^2) * dy = dx * (cos(x) - sin(x)/sqrt(2))
sqrt(1 - sin(x)^2 - 2sin(x)cos(x)/sqrt(2) - cos(x)^2 / 2) * dy = dx * (cos(x) - sin(x)/sqrt(2))
sqrt((1/2) * cos(x)^2 - sqrt(2) * sin(x)cos(x)) * dy = dx * (cos(x) - sin(x)/sqrt(2))
dy/dx = (cos(x) - sin(x)/sqrt(2)) / sqrt((1/2) * cos(x)^2 - sqrt(2) * sin(x)cos(x))
shreenikethan:
you could had used chain rule
which is simple
yes ...
I got my answer but you need to learn math first before giving advice.
Answered by
11
use identities of trigonometry
Attachments:
I got my answer but you need to learn math
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