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Title: Chain rule help Post by: bio_man on Sep 10, 2018 \(y=\sqrt{u^2-1}\)
\(u=4\sqrt{x}\) 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 ;D Here's the rationalized version \(y'=\frac{8}{\sqrt{16x-1}}\cdot \frac{\sqrt{16x-1}}{\sqrt{16x-1}}\rightarrow \ \frac{8\sqrt{16x-1}}{16x-1}\) |