Math Is Fun Forum

Hey, I know how to solve it now, too!
The keys are
a) the derivative of cos²(x) is -2cos(x)sin(x)
b)

Your question is so inspiring!  I used to think this kinda trig and derivative mix is my last-to-solve.

It's really great to have someone knowing how to solve hard integrals to join!

2x(x+7)-x(x+3) = x²-1
x(2x+14-x-3)=x²-1
11x=-1
x=-1/11

I've written the proof or the derivation in post 3, following Ricky's fomula, so you can check my procedure and simply duplicate it before showing the formula. And that will make your solution complete.

Compare y=f(ax[sub]2[/sub]) and y=f(x)
when the two y equals, usually the value in the bracket equals.
ax[sub]2[/sub]=x
thus x[sub]2[/sub]=x/a

CDF(median)=1/2
CDF(lower quatile)=1/4
CDF(upper quatile)=3/4
-you can solve the three using the expression of CDF yourself now:D

Yeah, your proof is restricted to the base nonnegative, while the proof by binomial expension is restricted to exponent natural numbers.

It depends on whether y is defined as nonnegtive or nonpositive. In either case, you can solve y explicitly and substitude y with the expression in the solution, and get a global solution on a certain domain.

Or, you already know the point (x[sub]0[/sub],y[sub]0[/sub]), and you want to find the tangent slope at it, you will easily get a local solution.

By the way,

d(a[sup]k[/sup])/dk = k a[sup]k-1[/sup]

(sin^2(2x))^2-That's just one way.

[sin²(2x)²]'=2sin²(2x) [sin²(2x)]'=2sin²(2x) 2sin(2x)[sin(2x)]'

=4sin³(2x)cos(2x)[2x]'=4sin³(2x)cos(2x)2=8sin³(2x)cos(2x)

you just need to factor m and n into products of prime numbers.
m= p[sub]1[/sub][sup]i1[/sup] p[sub]2[/sub][sup]i2[/sup]... p[sub]l[/sub][sup]il[/sup]
n= q[sub]1[/sub][sup]j1[/sup] q[sub]2[/sub][sup]j2[/sup] ... q[sub]k[/sub][sup]jk[/sup]
m²=...
n²= ...
compare the two sets above, latter will contain former.and then take square roots

Luck ain't even lucky
Got to make your own breaks

I have both bad news and good news for you.

bad news is-your theorem has already exist. in differential equation field, there is a similar formula called Euler's Method to evaluate f(x+a), given f'(x+t) 0<t<a and f(x).

Still Congratulations! You have your Own Thought and  Creativity.

Keep this thinking habit and Eventually you'll break a point in some field!