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

please show me how to do this!

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)

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