To determine

To calculate:

y'

Answer

y'= - cos⁡x-sin⁡x*(1-cos⁡x)sin2⁡(x-sin⁡x)

Explanation

Rule:

i. Reciprocal rule:

gx=1fx= - f'xfx2

ii. Chain rule:

dydx=dydu*dudx

iii.

ddx(sin x⁡)=cos⁡x

iv. Difference rule:

ddxfx-g(x)=ddxfx-ddxgx

Given:

y=1sin⁡(x-sin⁡x)

Calculation:

y= 1sin⁡(x-sin⁡x)

Using the reciprocal rule,

y'= - (sin⁡(x-sin⁡x))' [sin⁡(x-sin⁡x)]2

Using chain rule for numerator,

y'= - cos⁡x-sin⁡x*(x-sin⁡x)' [sin⁡(x-sin⁡x)]2

Using difference rule,

y'= - cos⁡x-sin⁡x*(1-cos⁡x)[sin⁡(x-sin⁡x)]2

Rewriting the denominator in the usual form,

y'= - cos⁡x-sin⁡x*(1-cos⁡x)sin2⁡(x-sin⁡x)

Conclusion:

y'= - cos⁡x-sin⁡x*(1-cos⁡x)sin2⁡(x-sin⁡x)