To determine
To find: The derivative of y
Answer
Answer:
y'=sec2x+secx-11+cosx
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(tanx)=sec2x and ddx(cosx)= -sinx
iv.
Constant rule, ddxc= 0
v.
tanx=sinxcosx, cosx=1secx,sin2x+cos2x=1
vi.
a2-b2=a+b(a-b)
Given:
y=tanx1+cosx
Calculation:
Consider the given function
y=tanx1+cosx
Differentiate with respect to x on both sides
ddxy=ddxtanx1+cosx
Applying the quotient rule of differentiation
y'=(1+cosx)*ddx(tanx)-tanx*ddx(1+cosx)(1+cosx )2
By applying the trigonometric and sum rule of differentiation
y'=(1+cosx)*sec2x-tanx ddx1+ddxcosx(1+cosx)2
y'= (1+cosx)sec2x-tanx (0-sinx)(1+cosx)2
y'= (1+cosx)sec2x+tanxsinx(1+cosx)2
y'=1+cosxsec2x+sinxcosx*sinx(1+cosx)2
By applying the trigonometric relations
y'=1+cosxsec2x+sin2x*secx(1+cosx)2
y'=1+cosxsec2x+(1-cos2x)*secx(1+cosx)2
y'=1+cosxsec2x+1+cosx1-cosxsecx(1+cosx)2
Factor out the common and simplifying
y'=1+cosx[sec2x+1-cosxsecx](1+cosx)2
y'=1+cosx[sec2x+secx-cosx secx](1+cosx)2
y'=sec2x+secx-11+cosx
Conclusion:
y'=sec2x+secx-11+cosx