AboutAlon Mandes Expertise Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems.
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Experience 1. I'm a team member of mathnerds (math site for answering questions)
2. I'm a team member in the Student's Union of the Technion, helping
students who have problems in mathematics.
3. 2 years of experience as a math teacher in college.
4. I give free homework help for high school students in
Mathematics & Physics.
5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
"Complex Functions".
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:
