use the first principle to obtain the differencial coefficient
y=1/(2x-1)
Try WolframAlpha :p
It's a free and really helpful math site :)
It should help you w / this question
how do i get into dat site u recommended
http://www.wolframalpha.com/
thank's Ms. Sue- I'm having trouble posting links ://
I know. Only a few of us tutors can post links.
not even sure what a "differencial coefficient" is
Do you mean to find the derivative, using the definition as a limit?
To obtain the differential coefficient of the given function y = 1/(2x-1) using the first principles, we need to compute the derivative from the definition of a derivative.
The first principles state that the derivative of a function is the limit of the difference quotient as the change in input approaches zero.
Let's proceed step by step:
1. Start with the definition of the derivative:
f'(x) = lim(h -> 0) [ f(x + h) - f(x) ] / h
Here, f(x) = 1/(2x-1).
2. Substitute the given function into the definition:
f'(x) = lim(h -> 0) [ 1/(2(x + h) - 1) - 1/(2x - 1) ] / h
3. Simplify the expression by finding a common denominator:
f'(x) = lim(h -> 0) [ (2x - 1) - (2(x + h) - 1) ] / [ h(2(x + h) - 1)(2x - 1) ]
4. Expand and simplify the numerator:
f'(x) = lim(h -> 0) [ 2x - 1 - 2x - 2h + 1 ] / [ h(2(x + h) - 1)(2x - 1) ]
= lim(h -> 0) [ -2h ] / [ h(2(x + h) - 1)(2x - 1) ]
5. Cancel out the common term 'h' in the numerator and denominator:
f'(x) = lim(h -> 0) [ -2 ] / [ (2(x + h) - 1)(2x - 1) ]
= -2 / [ (2x + 2h - 1)(2x - 1) ]
6. Take the limit as 'h' approaches zero:
f'(x) = -2 / [ (2x - 1)(2x - 1) ]
= -2 / [ 4x^2 - 4x + 1 ]
7. Simplify further if required:
f'(x) = -2 / (4x^2 - 4x + 1)
Therefore, the differential coefficient of y = 1/(2x-1) with respect to x is given by f'(x) = -2 / (4x^2 - 4x + 1).