I was a teacher in Srilanka from 1992 to 2006. I took tuition from 1990 to 1991 in India. I do general work in Canada but I help to do homework to my daughters. I finished my teacher's training college diploma in Srilanka. I completed my B.Sc degree course at Bharayiar university in India.
Bsc Mathematics India
Subjects of Expertise
SAT Math, ACT Math, GRE Math, GMAT Math, Pre-Algebra, Algebra 1, Algebra 2, Geometry, Trigonometry, Calculus, Statistics, Pre-Calculus, Elementary Math (K-6th)
|Last Login||Oct 26th 2020|
|Registered Since||Nov 27th 2017|
y=v(u(x)) — dy/dx=v'(u(x)).u'(x) (1) d/dx(sinx)=cosx (2) d/dx(cosx)=-sinx (3) d/dx(tanx)=(secx)2
In general, to find the derivative of a function defined parametrically by the equations x=u(t), y=v(t), we use the following rule dy/dx=(dy/dt).(dt/dx)=v(t)/u(t)
x=u(t) dx/dt=U'(t) y=v(t) dy/dt=v'(t) dy/dx=dy/dt.dt/dx d2y/dx2=d/dt(dy/dx).dt/dx
We can rewrite each rational term in the expression as a negative power of x
the formula for the derivative of arctan(x)=1/(1+x2).
In implicit differentiation, we differentiate both sides of the equation according to x and treat y as an implicit function of x