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Answer 1 · 2017-12-04T03:54:09

$\frac{cos (x)}{2 x - 1} - \frac{2 sin (x)}{{(2 x - 1)}^{2}}$ $color(blue)(cos(x)/(2x-1) - (2sin(x))/((2x-1)^2))$

We are given the function $f (x) = y = \frac{sin x}{2 x - 1}$ $f(x) = y = Sin x/(2x-1)$ $E x p r e s s i o n .1$ $color(red)(Expression.1)$

We need to find the First Derivative of $f (x)$ $f(x)$

By observing $f (x)$ $f(x)$ we know that we must use the Quotient Rule to differentiate .

Quotient Rule for finding the derivatives states that

$color(blue)((dy)/(dx)[f(x)/g(x)] = [(g(x)*f'(x) - f(x)*g'(x))/[g(x)]^2]$

Using the Quotient Rule, we can write our $color(red)(Expression.1)$ as

$(d/dx(sin x)(2x-1) - sin x(d/dx)(2x-1)]/(2x-1)^2$ $color(red)(Expression.2)$

We know that

$color(blue)(d/dx(sin x)$ is $color(green)(Cos x)$ $color(red)(..1)$

We can differentiate $color(blue)(2x-1)$ as follows:

$color(blue)(d/dx(2x - 1) =)$ $color(green)(2d/dxx + (d/dx)(-1)$ $color(red)(..2)$

We know that

$color(blue)(2d/dx(x) = 2*1)$ and $color(green)(d/dx(-1) = 0)$ $color(red)(..3)$

Hence,

using our intermediate results $color(red)(1,2, and 3)$ and our $color(red)(Expression.2)$

we can write

$color(blue)((Cos x*(2x - 1) - 2 * Sin x)/(2x-1)^2)$ $color(red)(..2)$

We can write $color(red)((2x-1)^2$ as $color(red)((2x-1)(2x-1))$

We can split the terms in $color(red)(..2)$ as follows and write our result:

$color(blue)((Cos x*(2x - 1))/((2x-1)(2x-1)) - (2*sin x)/((2x-1)(2x-1)$

We can now cancel common factors to simplify:

$color(blue)((Cos x*cancel(2x - 1))/(cancel(2x-1)(2x-1)) - (2*sin x)/((2x-1)(2x-1)$

$color(blue)(cos x/(2x-1) - (2*sinx)/((2x-1)^2))$

I hope this helps you to understand how the **Quotient Rule ** for differentiation works.