To determine

To find: The derivative of y

Answer

Answer:

y'=sec2⁡x+sec⁡x-11+cos⁡x

Explanation

Formula:

i.

Sum rule,  ddxfx+gx=ddx(fx+ddx(g(x)

ii.

Quotient rule,  ddxfxgx=gx*ddx(fx- fx*ddx(gxgx2

iii.

Trigonometric  rules,ddx(tan⁡x)=sec2⁡x  and ddx(cos⁡x)= -sin⁡x

iv.

Constant rule, ddxc= 0

v.

tanx=sinxcosx, cosx=1secx,sin2⁡x+cos2⁡x=1

vi.

a2-b2=a+b(a-b)

Given:

y=tan⁡x1+cos⁡x

Calculation:

Consider the given function

y=tan⁡x1+cos⁡x

Differentiate with respect to x on both sides

ddxy=ddxtan⁡x1+cos⁡x

Applying the quotient rule of differentiation

y'=(1+cos⁡x)*ddx(tan⁡x)-tan⁡x*ddx(1+cos⁡x)(1+cos⁡x )2

By applying the trigonometric and sum rule of differentiation

y'=(1+cos⁡x)*sec2⁡x-tan⁡x ddx1+ddxcosx(1+cos⁡x)2

y'= (1+cos⁡x)sec2⁡x-tan⁡x (0-sin⁡x)(1+cos⁡x)2

y'= (1+cos⁡x)sec2⁡x+tan⁡xsin⁡x(1+cos⁡x)2

y'=1+cosxsec2⁡x+sinxcosx*sinx(1+cos⁡x)2

By applying the trigonometric relations

y'=1+cosxsec2⁡x+sin2⁡x*secx(1+cos⁡x)2

y'=1+cosxsec2⁡x+(1-cos2⁡x)*secx(1+cos⁡x)2

y'=1+cosxsec2⁡x+1+cosx1-cosxsecx(1+cos⁡x)2

Factor out the common and simplifying

y'=1+cosx[sec2⁡x+1-cosxsecx](1+cos⁡x)2

y'=1+cosx[sec2⁡x+secx-cosx secx](1+cos⁡x)2

y'=sec2⁡x+sec⁡x-11+cos⁡x

Conclusion:

y'=sec2⁡x+sec⁡x-11+cos⁡x