Question

Find the differential $dy$. $y=\frac{x+1}{2x-1}$

Short answer

The differential $dy$ of the function $y=\frac{x+1}{2x-1}$ is $-\frac{3}{(2x-1{)}^2}\text{.}dx$.

Step-by-step solution

Identify $u$ and $v$

In our function $y=\frac{x+1}{2x-1}$, $u$ can be identified as $x+1$ and $v$ as $2x-1$.

Compute the Derivatives $u^{\prime }$ and $v^{\prime }$

Compute the derivative of $u$ with respect to $x$ to get $u^{\prime }=1$. Next, compute the derivative of $v$ with respect to $x$ to get $v^{\prime }=2$.

Apply the Quotient Rule

Now that we have $u$, $v$, $u^{\prime }$, and $v^{\prime }$, we can apply the quotient rule $\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$ which grants us $\frac{(2x-1)(1)-(x+1)(2)}{(2x-1{)}^2}$. This simplifies to $\frac{-3}{(2x-1{)}^2}$.

Compute the Differential $dy$

To find the differential $dy$, we just have to multiply the derivative by $dx$. So, $dy=-\frac{3}{(2x-1{)}^2}\text{.}dx$.