Expert: NUR ARINA BASILAH BINTI KAMISAN - 7/24/2008

Question
Thanks for helping me in answering my questions, may you please help me with the following?

Determine dy/dx and simplify where possible:
1. y = x^5 sin x
2. y =ln(x^square root 1-x^2)
3. y = e^sin25x
4. y = tan(x^-3)
5. y = 4^3x-1
6. y =1/square root cossquare root x
7. y =(cotx-3secx)^1/2

Answer
dear betty,

1.you have to differentate it part by part
y=x^5sinx
dy/dx=(u*dv/dx)+v*du/dx)
let u=x^5 --> du/dx=5x^4
and v=sinx --> dv/dx=cosx
so dy/dx=(x^5)*(cosx)+(sinx)*(5x^4)
       =x^5cosx+5x^4sinx

2.y=ln(x^square root 1-x^2)
bring the power of x to the front of equation (lnx^5=5lnx) and you wll have
y=(square root 1-x^2)lnx
by using method in the first question you will have
u=square root 1-x^2, du/dx=-x(1-x^2)^-1/2
v=ln x, dv/dx=1/x
so dy/dx=(square root 1-x^2)/x+(-x)(1-x^2)^-1/2lnx

3.y=e(sin^2 5x)
and (sin^2 5x=1-cos5x/2)-->(from sin^2 x=1-cosx/2)-->identity of trigonometry
y=e(1-cos5x/2)
dy/dx=e(1-cos5x/2)*sin5x/2*5
    =(5sin5x/2)e(1-cos5x/2)

4.y=tanx^-3
dy/dx=sec^2 (x^-3)*(-3x^-4)
     =(-3x^-4)sec^2(x^-3)

5.y=4^3x-1 (3^x+y=3^x*3^y)
so the equation will become:
y=4^3x/4
y=16^x (since 4^3/4=16)
dy/dx=ln16(16^x) -->(a^x,dy/dx=lna(a^x))

6.y=1/square root cos(square root x)
y=(cos(square root x))^-1/2
dy/dx=-1/2(cos (square root x))^-3/2*(-sin(square root x))*(1/2x^^-1/2)

7.y=(cotx-3secx)^1/2
dy/dx=1/2(cotx-3secx)^-1/2*(-cosec^2(x)-3secxtanx)

note that if you have a question like number 3,4,6 and 7, you have to differentiate them one by one.you can refer note on this at this website http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/trigderivdirectory/TrigDeriv...

good luck!