First step: partition the number you want to square root into a block of 2 digits, starting from the last digit (first diagram)

Second step: As our number is a five-digits, we ends up with 2 28 01. Pick a number that could be squared to get the first partition, 2. This number is 1, since 1×1=1

Third step: Write 1 on the top and on the side, as shown in the second graph

Fourth step: Double the number on the side, which is 1+1=2 and use this number as the first digit for the next multiplier. Meanwhile, subtract 1 from 2 inside the root sign to get 1, then pull the other two digits, 28

Fifth step: We need a value in the boxes that when we multiply together will give a number less than 128. We choose 5 as 25×5=125

Sixth step: Subtract 125 from 128 to give 3, and as the same concept with long division, bring down the 0 and the 1. So we have 301

Seventh step: Add 5 to the multiplier on the left, so 20+5=30, which we will use on the side as the hundred and ten digits.

Final step: Find a number to fit in the boxes. We choose 1 since 301×1=301

And hence the square root of 228801 is 151

3 0

2 years ago

F(x) = 4√(2x³-1)<br> F'(x) =.....?

goldfiish [28.3K]

Answer:

$\large\boxed{f'(x)=\dfrac{12x^2}{\sqrt{2x^3-1}}}$

Step-by-step explanation:

$f(x)=4\sqrt{2x^3-1}=4\left(2x^3-1\right)^\frac{1}{2}\\\\f'(x)=4\cdot\dfrac{1}{2}(2x^3-1)^{-\frac{1}{2}}\cdot3\cdot2x^2=\dfrac{12x^2}{(2x^3-1)^\frac{1}{2}}=\dfrac{12x^2}{\sqrt{2x^3-1}}\\\\\text{used}\\\\\sqrt{a}=a^\frac{1}{2}\\\\\bigg[f\left(g(x)\right)\bigg]'=f'(g(x))\cdot g'(x)\\\\\bigg[nf(x)\bigg]'=nf'(x)\\\\(x^n)'=nx^{n-1}$