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Chain rule help
bio_man
bio_man
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4 months ago
Chain rule help
\(y=\sqrt{u^2-1}\)
\(u=4\sqrt{x}\)
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
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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