Calculus: dx/dy = nax^(n-1). Can someone tell me the theory and the Proof?

Question

dx = anx(n-1)
dy
Where a is the coefficient (a>0)
N is the power.

Answer 1

It's based upon the fact that the derivative IS the slope of a line tangent to a curve.

You need two points to calculate slope, m = (y2-y1)/(x2-x1)

But

Tangent means the line touches the curve at exactly one point.

So, theoretically, you will find the slope using 2 x-values that are "h" amount of space apart from each other.

so, the points you are work with are (x, f(x)) and (x+h, f(x+h))

when you go to find the slope, you'd calculate:

f(x + h) - f(x)
----------------------, OR
(x + h) - x


f(x + h) - f(x)
-----------------------, HOWEVER...
h

Since we are looking for the slope of the line TANGENT to a curve, we want the distance between the x-values to be zero.

Hence, the magic formula...

f'(x) = lim h ->0 f( x+ h) - f(x)
.......................... ---------------------
...................................... h

After playing around with this formula, a pattern develops. That pattern is, THE POWER RULE.

Don't believe me? Check it out:

f(x) = x^2

so f(x + h) = (x + h)^2
..................= x^2 +2xh + h^2

So, f'(x) = lim h->0 [x^2 +2xh + h^2] - x^2
...................................... -----------------------------------
........................................................... h

................= lim h->0 2xh + h^2
...................................... ---------------------
.............................................. h

................= lim h->0 2x + h

................= 2x

See, it's a piece of cake! :)

Answer 2

I'm really not sure what you're asking. Are you sure it's dx/dy and not dy/dx?

Possibility 1 - you want the proof of the rule that says if y = ax^n then dy/dx = anx^(n-1). This comes from the definition of the derivative:
lim(h->0) (y(x+h)-y(x)) / h
= lim(h->0) (a(x+h)^n - ax^n) / h
= lim(h->0) (ax^n + anx^(n-1)h + a(n 2)x^(n-2)h^2 + ... + ah^n - ax^n) / h
= lim(h->0) anx^(n-1) + a(n 2)x^(n-2)h + ... + ah^(n-1)
= anx^(n-1) + 0 + ... + 0
= anx^(n-1).

Possibility 2: Solve dy/dx = anx^(n-1). As above, the solution is y = ax^n.

Possibility 3: Solve dx/dy = anx^(n-1). Then dy/dx = 1/(anx^(n-1)) = x^(1-n)/(an), so y = x^(2-n)/(an(2-n)) for n ≠ 2; if n = 2 then dy/dx = 1/(2ax), so y = ln |x| / 2a.

Answer 3

Let y = x^n
dy/dx = lim delta x-> 0 ((x+delta x)^n - x^n)/(delta x)

Use the binominal expansion of (x+delta x)^n

The first term will be x^n which cancels out with the other original term in the numerator, -x^n

The second term in the binominal expansion
is n x^(n-1) (delta x)
all other terms have higher powers of (delta x)

factor the one delta x in the numerator and cancel with the denominator

Then when you take the limit all terms except for n x^(n-1) become 0.

you can then solve using a similar principle, that if derivative of f(x) is y', then derivative of (a)f(x) is ay'

Answer 4

an example of that would be if u had 4x^3...you would subtract one from the exponent and take the exponent (3) and put that in the front of the problem with a multiplication sign..so u would have (3) x 4x^2 = 12x^2