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

2 Answers
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)

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