0
√(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)]
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What is the derivative of square root of 5x?
Derivative with respect to 'x' of (5x)1/2 = (1/2) (5x)-1/2 (5) = 2.5/sqrt(5x)
What is the derivative of square root 2x-x2?
0
What is the derivative of lnx?
Derivative of lnx= (1/x)*(derivative of x) example: Find derivative of ln2x d(ln2x)/dx = (1/2x)*d(2x)/dx = (1/2x)*2===>1/x When the problem is like ln2x^2 or ln-square root of x...., the answer won't come out in form of 1/x.
What is a principal square root?
The principal square root is the non-negative square root.
Simplify square root of 5 Times Square root of 6?
To simplify the square root of 5 times the square root of 6, you can multiply the two square roots together. This gives you the square root of (5*6), which simplifies to the square root of 30. Therefore, the simplified answer is the square root of 30.
