wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find the derivative of cosec2x, by using first principle of derivatives ?

Open in App
Solution

y=csc2x .......(1)
Increase from y to y+δy correspondingly x to x+δx in the above equation(1)
y+δy=csc2(x+δx) .....(2)
Eqn(2)-Eqn(1)
y+δyy=csc2(x+δx)csc2x
δy=1sin2(x+δx)1sin2x
δy=sin2xsin2(x+δx)sin2xsin2(x+δx)
δ=sin(x+x+δx)sin(xxδx)sin2xsin2(x+δx) using sin2Asin2B=sin(A+B)sin(AB)
δy=sin(x+x+δx)sin(xxδx)sin2xsin2(x+δx)
δy=sin(2x+δx)sin(δx)sin2xsin2(x+δx)
δy=sin(2x+δx)sin(δx)sin2xsin2(x+δx) since sin(θ)=sinθ
limδx0δyδx=limδx0sin(2x+δx)sinδxδxsin2xsin2(x+δx) by applying limits to both sides
dydx=sin(2x+0)sinδxδxsin2xsin2(x+0)
dydx=sin2xsin2xsin2x since limθ0sinθθ=1
dydx=2sinxcosxsin2xsin2x since sin2x=2sinxcosx
dydx=2cosxsin3x
dydx=2cotxcsc2x

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Derivative of Standard Functions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon