About NUR ARINA BASILAH BINTI KAMISAN Expertise i can answer derivatives and integral questions but not so expert in answering limits,vector and infinite series.
Experience i teached calculus for foundation students at university malaya and i also teached tutorial class in university for subject differential equation when i was doing my degree.currently i'm teaching algebra for foundation students.
Education/Credentials i have degree in mathematics with education and now i'm futhering my study in master of statistics at university malaya.
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)
