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Posts: 23128
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$$y'=\frac{8}{\sqrt{16x-1}}\cdot \frac{\sqrt{16x-1}}{\sqrt{16x-1}}\rightarrow \ \frac{8\sqrt{16x-1}}{16x-1}$$ Read 168 times
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