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Posts: 23921
6 months ago
\(y=\sqrt{u^2-1}\)
\(u=4\sqrt{x}\)

Put u into x

\(y=\sqrt{\left(4\sqrt{x}\right)^2-1}\)

Simplify:

\(y=\sqrt{16x-1}\)

Now derive, change to fractional coefficients:

\(y=\left(16x-1\right)^{\frac{1}{2}}\)

Apply the chain rule (same as power rule by subtracting the exponent by 1, multiplying by the exponent as a factor, then multiplying by the derivative of what's inside the brackets)

\(y=\frac{1}{2}\left(16x-1\right)^{\left(\frac{1}{2}-1\right)}\times \left(\frac{d}{dx}\left(16x-1\right)\right)\)

Let's find the derivative of 16x minus 1:

\(y'=\frac{1}{2}\left(16x-1\right)^{\left(-\frac{1}{2}\right)}\times 16\)

Derivation is done, now simplify more:

\(y'=\frac{16\left(16x-1\right)^{\left(-\frac{1}{2}\right)}}{2}\)

Some more:

\(y'=8\left(16x-1\right)^{\left(-\frac{1}{2}\right)}\)

Negative exponents are ugly, so use negative exponent rule of exponents and then change to radical form:

\(y'=\frac{8}{\left(16x-1\right)^{\frac{1}{2}}}\rightarrow \frac{8}{\sqrt{16x-1}}\)

You're done, if you want to rationalize the denominator, you can do that too, it's an extra step Grinning Face

Here's the rationalized version

\(y'=\frac{8}{\sqrt{16x-1}}\cdot \frac{\sqrt{16x-1}}{\sqrt{16x-1}}\rightarrow \ \frac{8\sqrt{16x-1}}{16x-1}\)
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