Show d/dx (cot x)=-(cosec x)^2.?

2 Answers
Øko
Mar 5, 2018

See explanation

Explanation:

We want find the derivative of

#y=cot(x)=cos(x)/sin(x)#

Use the quotient rule, if #y=f/g#

then #dy/dx=(f'*g-f*g')/g^2#, thus

  • #f=cos(x)=>f'=-sin(x)#
  • #g=sin(x)=>g'=cos(x)#

Thus

#dy/dx=(-sin(x)*sin(x)-cos(x)cos(x))/sin^2(x)#

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

#=-1/sin^2(x)=-csc^2(x)#

Or

#d/dxcot(x)=-csc^2(x)#

Jim G.
Mar 5, 2018

#"see explanation"#

Explanation:

#"differentiate using the "color(blue)"quotient rule"#

#"given "f(x)=(g(x))/(h(x))" then"#

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

#f(x)=cotx=cosx/sinx#

#g(x)=cosxrArrg'(x)=-sinx#

#h(x)=sinxrArrh'(x)=cosx#

#rArrd/dx(cotx)#

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

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

#=-1/sin^2x=-csc^2x#

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