i believe these questions are going to only resu

Its true that that is a good method and probably the traditional one they teach in school but, its not optimal.

you only asked for a solution, not an optimal solution, I believe BB's solution is equally on par with yours(if not better imo).

I was expecting your questions to be a bit different then what you posted, as these arn't really math "tricks" as much as math tips.

also, for your first question, in high school I became very obsessed with discovering that formula, I was always trying to figure out another way to devise the square of a number instead of multiplying the number, too which i eventually figured out the formula of n^2 =(n-1)^2+( (n-1)^2-(n-2)^2+2); or counting a delta upwards by 2 to be added to each squared number to reach the next.

my turn
This is one i use to catch out 18-21 year old maths students, so lets see how gamedev does.

Evaluate d/dx of 2[sup]x[/sup]

I love math. I've got a trick for multiplying by 9 which does probably come up more often than by 999999

First round up to the nearest 10s place and then divide by 10 to get the magic number.
0 - 10 rounds up to 10, becomes 1,
11 - 20 rounds up to 20, becomes 2,
21 - 30 rounds up to 30, becomes 3, etc...

Subtract the magic number from the original number and then add a digit onto the end which will make all of the resulting digits add up to 9. Some examples.

9 * 13

1. magic number is 2
   
2. 13 - 2 = 11
   
3. 1 + 1 + x = 9 : x = 7
   
4. And the answer is 117.
   

9 * 57

1. magic number is 6
   
2. 57 - 6 = 51
   
3. 5 + 1 + x = 9 : x = 3
   
4. and you get 513
   

my turn
This is one i use to catch out 18-21 year old maths students, so lets see how gamedev does.

Evaluate d/dx of 2[sup]x[/sup]

perhaps i'm missing sometin, but "d/dx" = x...right?

perhaps i'm missing sometin, but "d/dx" = x...right?

Don't you mean 1/x? Coz the d's cancel out

[quote name='slicer4ever' timestamp='1346823734' post='4976704']
perhaps i'm missing sometin, but "d/dx" = x...right?

Don't you mean 1/x? Coz the d's cancel out
[/quote]
it'd be 1*x, since dx is multiplying d*x, and by order of op's it'd expand to: (d/d)*x

...maybe....?

it's be 1*x, since dx is multiplying d*x, and by order of op's it'd expand to: (d/d)*x

...maybe....?

Oh you would be right. But then what purpose does 2^x serve? Perhaps we need to multiply it.... my answer is x * 2^x.

And that's 3^x since there's one more x. I think, anyway - my math is a bit rusty!

Although seriously, while the answer to d/dx 2^x is fairly straightforward if you know the formula for differentiating exponentials, it is interesting to derive it yourself.

I am going to assume d/dx means "take the derivative with respect to x". In that case it is log2*2^x. This is because 2^x is equal to e^(x*log2) (where log is the natural logarithm i.e. in base e) so since the derivative of e^ax with respect to x is a*e^ax, we get log2*e^(x*log2) which is equal to the result given above.
With the squares, you can exploit the fact that the difference between n^2 and (n+1)^2 is 2n+1 (using the difference of two squares formula), so you can use addition to get the squares of the first 50 counting numbers.