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

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please show me how to do this!

Everardo Feil · Accepted Answer

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 xSo 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)^2Split 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 xNow you have that -csc x cot x = -csc x cot x.Hope this helps!

Everardo Feil · Answer

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)

Everardo Feil · Answer

D Dx Csc

Everardo Feil · Answer

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)

Everardo Feil · Answer

Derivative of u/v = (u’v-uv’)/v^2Derivative 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)xNevermind, this is totally wrong.

Everardo Feil · Answer

d/dx(cscx)=d/dx(1/sinx)quotient rule:d/dx(1/sinx)=-cosx/sin^2x=-(cosx/sinx)(1/sinx)=-(cotx)(cscx)

Oran Armstrong

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

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