To determine

To calculate:

y'

Answer

y'=-sec2⁡1-x21-x

Explanation

Rule:

i. Power rule combined with chain rule:

ddxun=nun-1 .ddx(u)

ii. Difference rule:

ddxfx-g(x)=ddxfx-ddxgx

iii.

ddx(tan⁡x)=sec2⁡x

Given:

y=tan⁡(1-x )

Calculation:

y= tan⁡(1-x )

Let y=tan⁡u

Where u=1-x

Differentiate y with respect to x

y'=ddxtan⁡(1-x )

By using chain rule

y'=ddutan⁡u.ddx(u)

By using ddx(tan⁡x)=sec2⁡x

y'=sec2⁡u.ddx(u)

Substitute value of u

y'=sec2⁡1-x.ddx(1-x)

By using power rule combined with chain rule

y'=sec2⁡1-x.121-xddx(1-x)

Since by using power rule ddxx=12x

Now by using Difference rule

y'=sec2⁡1-x.121-xddx1-ddxx

Derivative of constant is zero and by using power rule

y'=sec2⁡1-x.121-x0-1

y'=sec2⁡1-x.-121-x

By rearrangement

y'=-sec2⁡1-x21-x

Conclusion:

y'=-sec2⁡1-x21-x