Question

how to align at equal signs ?

Answer 1

If I have a list of steps like this, how can I make the equal sign aligned?, and how can I prevent the extra space between steps ?, I tried to use inline \ ( \ ) instead of newline \ [ \ ], It did reduce the space, but it also made the font really tiny

Thanks

\(y = x^{\frac{1}{n}}\)
\(y^n = x\)
\[y' = \frac{1}{n} y^{(1-n)}\]
\[= \frac{1}{n} \frac{y^1}{y^{n}} \]
\[= \frac{1}{n} \frac{x^{\frac{1}{n}}}{y^{n}} \]
\[= \frac{1}{n} \frac{x^{\frac{1}{n}}}{(x^{\frac{1}{n}})^{n}} \]
\[= \frac{1}{n} \frac{x^{\frac{1}{n}}}{x} \]
\[= \frac{1}{n} x^{\frac{1}{n}}{x^{(-1)}} \]
\[= \frac{1}{n} x^{\frac{1}{n} - 1 }\]

Answer 2

\begin{eqnarray*}
y =& &x^{\frac{1}{n}} \\
y^n =& &x \\
y' =& &\frac{1}{n} y^{(1-n)} \\
=& &\frac{1}{n} \frac{y^1}{y^{n}} \\
=& &\frac{1}{n} \frac{x^{\frac{1}{n}}}{(x^{\frac{1}{n}})^{n}} \\
=& &\frac{1}{n} \frac{x^{\frac{1}{n}}}{x} \\
=& &\frac{1}{n} x^{\frac{1}{n}}{x^{(-1)}} \\
=& &\frac{1}{n} x^{\frac{1}{n} - 1 }\\
\end{eqnarray*}

Answer 3

I think I found it
\ begin{eqnarray*}
\ end{eqnarray*}

Answer 4

```
\[\begin{array}{rl}
y &=x^{\frac1n} \\
y^n &=x \\
y' &=\frac1n y^{(1-n)} \\
&=\frac1n\frac{y^1}{y^n} \\
&=\frac1n \frac{x^{\frac1n}}{(x^{\frac1n})^n} \\
&= \frac1n\frac{x^{\frac1n}}x \\
&= \frac1nx^{\frac1n}{x^{(-1)}} \\
&= \frac1n x^{{\frac1n}- 1}
\end{array}\]
```
\[\begin{array}{rl}
y &=x^{\frac1n} \\
y^n &=x \\
y' &=\frac1n y^{(1-n)} \\
&=\frac1n\frac{y^1}{y^n} \\
&=\frac1n \frac{x^{\frac1n}}{(x^{\frac1n})^n} \\
&= \frac1n\frac{x^{\frac1n}}x \\
&= \frac1nx^{\frac1n}{x^{(-1)}} \\
&= \frac1n x^{{\frac1n}- 1}
\end{array}\]

Answer 5

or simply



```
\[\begin{align}
y &=x^{\frac1n} \\
y^n &=x \\
y' &=\frac1n y^{(1-n)} \\
&=\frac1n\frac{y^1}{y^n} \\
&=\frac1n \frac{x^{\frac1n}}{(x^{\frac1n})^n} \\
&= \frac1n\frac{x^{\frac1n}}x \\
&= \frac1nx^{\frac1n}{x^{(-1)}} \\
&= \frac1n x^{{\frac1n}- 1}
\end{align}\]
```

\[\begin{align}
y &=x^{\frac1n} \\
y^n &=x \\
y' &=\frac1n y^{(1-n)} \\
&=\frac1n\frac{y^1}{y^n} \\
&=\frac1n \frac{x^{\frac1n}}{(x^{\frac1n})^n} \\
&= \frac1n\frac{x^{\frac1n}}x \\
&= \frac1nx^{\frac1n}{x^{(-1)}} \\
&= \frac1n x^{{\frac1n}- 1}
\end{align}\]