What is the derivative of the square root of 1-sinx?

Answer 1

√(1-sinx)=(1-sinx)1/2

Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)

d/dx(1-sinx)1/2=(1/2)(1-sinx)1/2-1*d/dx(1-sinx)

d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*d/dx(1-sinx)

-The derivative of 1-sinx is:

d/dx(u-v)=du/dx-dv/dx

d/dx(1-sinx)=d/dx(1)-d/dx(sinx)

d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[d/dx(1)-d/dx(sinx)]

-The derivative of 1 is 0 because it is a constant.

-The derivative of sinx is:

d/dx(sinu)=cos(u)*d/dx(u)

d/dx(sinx)=cos(x)*d/dx(x)

d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*d/dx(x))]

-The derivative of x is:

d/dx(xn)=nxn-1

d/dx(x)=1*x1-1

d/dx(x)=1*x0

d/dx(x)=1*(1)

d/dx(x)=1

d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*1)]

d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x))]

d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[-cos(x)]

d/dx(1-sinx)1/2=(-cosx)/[2√(1-sinx)]