Please prove this: d/dx (csc x) = – csc x cot x ?

6 Answers

• d/dx (csc x) = -csc x cot x

  You know that csc(x) = 1/sin(x), so try to make the left side of the equation look like the right side.

  d/dx (1/sin(x)) = -csc x cot x

  So from here just focused on manipulating left side.

  Use the quotient rule to find the derivative of 1/sin(x):

  [0 - 1*cos x] / [(sin x)^2] = -cos x / (sin x)^2

  Split up the (sin x)^2 term into sin x * sin x and rearrange fraction:

  [(cos x)/(sin x)] * (-1/sin x)

  You know that cos x / sin x = cot x & -1/sin x = -csc x

  Now you have that -csc x cot x = -csc x cot x.

  Hope this helps!

• OK check it:

  csc (x) = 1/ sin(x)

  d/dx 1/sin(x)= [sin(x)(0) - cos(x)(1)]/[sin(x)^2]=

  -cos(x)/(sin(x)^2)=

  -cos(x)/sin(x) * 1/sin(x)=

  -cot(x)csc(x)= -csc(x)cot(x)

  therefore:

  d/dx csc(x)= -csc(x)cot(x)

• D Dx Csc

• I'm assuming you have knowledge of the power rule and chain rule.

  f(x) = csc(x)

  f(x) = 1/sin(x)

  f(x) = [sin(x)]^(-1)

  Take the derivative and use the power rule and chain rule.

  f'(x) = (-1)[sin(x)]^(-2) [ cos(x) ]

  f'(x) = (-cos(x)) / sin^2(x)

  Split into two fractions,

  f'(x) = [-cos(x)/sin(x)] [ 1/sin(x) ]

  f'(x) = [ -cot(x) ] [ csc(x) ]

  f'(x) = -csc(x)cot(x)

• Derivative of u/v = (u'v-uv')/v^2

  Derivative of cos^(0.5) x = 0.5 x cos^(-0.5)x . (- sin x)

  The derivative of root(cos x/sin x) = (0.5cos^(-0.5)x . (- sin x) . sin^(0.5)x - cos^(0.5)x . 0.5sin^(-0.5)x . cos x) / sin x

  = - 0.5sin^(-0.5)x . cos^(-0.5)x (sin^2 x + cos^2 x) / sin x

  = - 0.5 / (sin^(1.5)x . cos^(0.5)x)

  = - 0.5 sin^(0.5)x / (sin^2 x cos^(0.5)x)

  = - 0.5 . csc^2 x . cot^(-0.5)x

  Nevermind, this is totally wrong.

• d/dx(cscx)=d/dx(1/sinx)

  quotient rule:

  d/dx(1/sinx)=-cosx/sin^2x

  =-(cosx/sinx)(1/sinx)

  =-(cotx)(cscx)