Expert: Alon Mandes - 8/27/2008

Question
what will be the derivatives of these functions?a) y=(4x+3)^4(x+1)^-3 ,b)f(theta)=(sin(theta)/1+cos(theta))^2 ,c)r=sec sqrt(theta)tan(1/theta), d)q=sin(t/sqrt t+1).i need answer of each question in detail, thank you very much.

Answer
a) y=[(4x+3)^4]*[(x+1)^(-3)]. Applying the rule of (fg)'=f'g+fg' :
  y'=[(4x+3)^4]'*[(x+1)^(-3)]+[(4x+3)^4]*[(x+1)^(-3)]'
  y'=[4*4*(4x+3)^3]*[(x+1)^(-3)]+[(4x+3)^4]*[-3*(x+1)^(-4)].
  y'=[16(4x+3)^3]*[(x+1)^(-3)]-3*[(4x+3)^4]*[(x+1)^(-4)].
  y'=[(4x+3)^3]*[(x+1)^(-3)]* { 16 - 3*[(4x+3)]*[(x+1)^(-1)] }.
  y'=[(4x+3)^3]*[(x+1)^(-3)]* { 16 - 3*(4x+3)/(x+1) }.

b) f(Ø)=( sin{Ø/[1+cos(Ø)]} )^2. Applying the rule :
  { [sin(g(x))]^2 }' = g'(x)*2*sin(g(x))*cos(g(x)) :
  f'(Ø)={Ø/[1+cos(Ø)]}' *2*sin{Ø/[1+cos(Ø)]}*cos{Ø/[1+cos(Ø)]}
  Let's calculate the derivative of the form Ø/[1+cos(Ø)]
  (Ø/[1+cosØ])'=[1(1+cosØ)- Ø*(-sinØ)]/[1+cosØ]^2. This was
  achieved according to the rule : (f/g)'=(f'g-g'f)/g^2. So,
  f'(Ø)={[1+cosØ+ØsinØ]/[1+cosØ]^2}*sin(2Ø/[1+cosØ])
  Explanation :
  We followed the trigonometric rule : 2sinxcosx=sin(2x).

c) r(Ө)=sec{sqrt(Ө)}*tan(1/Ө)
  Appling the rule : (fg)'=f'g+fg' , we get
  r'(Ө)=sec{sqrt(Ө)}'*tan(1/Ө)+sec{sqrt(Ө)}*tan(1/Ө)'
  r'(Ө)=[1/2sqrt(Ө)]*(sec{sqrt(Ө)}*tan{sqrt(Ө)}*tan(1/Ө) +
        [-1/Ө^2]*sec{sqrt(Ө)}*[sec(1/Ө)]^2
  Explanation :
  1. sec(f(x))'=f'(x)*sec(f(x))*tan(f(x))
  2. tan(g(x))'=g'(x)*[sec(g(x))]^2
  So,
  
d) q(t)=sin(t/sqrt t+1).Applying the rule:sin(f(x))'=f'(x)*cos(x)
  q'(t)=(t/sqrt t+1)' * cos(t/sqrt t+1)
  I didn't understand if you meant t/[sqrt(t)+1] or t/sqrt(t+1)
  So I'm going to calculate both & you take the one that
  you meant. In both cases we will apply the rule :
  (f/g)'=(f'g-g'f)/g^2. So, let's get to work:

  1. { t/[sqrt(t)+1] }' = { 1*[sqrt(t)+1]-t*[1/2sqrt(t)] } /
                          { [sqrt(t)+1]^2 }
                          

  2. { t/sqrt(t+1) } ' = {1*sqrt(t+)-t*1/[2sqrt(t+1)] } / {t+1}



I hope this will help you.
Alon.