Differentiation?

Question

Hi All,
Why is the derivative of the following
A(x)=(1-a/n+xa/n)^n
this
A'(x)=a(1-a/n+xa/n)^n-1
I am unsure about the differentiation of the terms inside the bracket. If this bracket was expanded surely there would be lots of terms with x in.
I understand where the a comes from, but not really why. the n comes down and is then cancelled out by the a/n.

Answer 1

Power Rule: (x^n)' = n*x^(n-1)
Constant Rule: (cx)' = c

We can apply power rule and chain rule to the following:
[(c+x)^n]'
= n*(c+x)^(n-1)*(c+x)'
= n*(c+x)^(n-1)*(0+1)
= n*(c+x)^(n-1)

Now suppose a and n are constants.
Then we can apply the above concepts to get:
[(c+(a/n)x)^n]'
= n*(c+(a/n)x)^(n-1)*(c+(a/n)x)'
= n*(c+(a/n)x)^(n-1)*(0+(a/n))
= n*(c+x)^(n-1)*(a/n)
= a*(c+x)^(n-1)

If we let c = 1 - a/n, we get what we need:
[(1-(a/n)+(a/n)x)^n]' = a*(1-(a/n)+x)^(n-1)

Answer 2

I presume you mean

A(x) = [1 - (a/n) + (xa/n)]^n

You were on the right track but didn't apply the chain rule. Keep in mind that we're dealing with almost all constants; the only nonconstant is (xa/n).

A'(x) = n[1 - (a/n) + (xa/n)]^(n - 1) {a/n}

(I denoted the chained stuff in { } brackets.)

The n and (a/n) multiplied together will give just a, so

A'(x) = a[1 - (a/n) + (xa/n)]^(n - 1)

Answer 3

Can you clear up what your differentiating with respec too? I'm assuming x is your variable and (a) a contant and (n) a real limit.

This looks like a taylor series approximation to me. Remember the tools of differentiation:

Product rule

Quotient Rule

Chain rule

Using these you should be able to solve easily.

Ah.. yes. Straight forward.

Try this.

For this we need the chain rule only. So we have a function A(x) which we are trying to differentiate with respect to (x). So we want:

dA(x)/dx = dA(x)/du * du/dx

Let u=(1-a/n+xa/n).

So du/dx=a/n --> since constants differentiate to zero.

Now back to our function:

A(x) = u^n

Therefore:

dA(x)/du=n.u^(n-1)

Now simply run the chain rule.

dA(x)/dx = n.u(n-1) * a/n

The (n) term cancels and leaves you with the answer you seek. simply resubstitute the u back in.

Answer 4

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Answer 5

The method that I got taught, that rarely ever fails for this kinda thing.

1. Drop the power in front.
2. Differentiate the bracket.
3. Drop the power by one.

So, applying these steps to A(x) = (1 - (a/n) + (a/n)x)^n...

1. Drop the power in front

n [1 - (a/n) + (a/n)x ]

2. Differentiate the bracket

[1 - (a/n) + (a/n)x ] => (a/n)

3. Drop the power by one

n => n-1

All of this goes to

n . (a/n) . [1 - (a/n) + (a/n)x ] ^ (n-1)

= a [1 - (a/n) + (a/n)x]^(n-1)

Answer 6

They are all rules. If you follow them correctly you get a correct answer.

A'(x) = a(1- a/n + xa/n)^(n-1)

And remember some imp. things that always take constants aside. And if you get a very bad looking expression, assume some repeated terms as y then differentiate the given expn wrt y, differentiate y wrt x (whatever's given) and multiply those. This is the chain rule.

dy/dx = dy/dz * dz/dx

Answer 7

A(x)=(1-a/n+xa/n)^n

Let X = (1-a/n+xa/n).

Then A(X )=X^n.

A’( X) = n X^(n-1).

A(x) = A’( X) d X / dx.

X = (1-a/n+xa/n).

d X / dx = a/n. since other terms are constants.

A(x) = A’(X) d X / dx

=n X^(n-1) * a/n.

= a X^(n-1)

Replace X by (1-a/n+xa/n)

Answer 8

Use the chain rule - dA/dx = dA/dv x dv/dx,
where v = 1 - a/n + xa/n, so A = v^n

dA/dv = nv^(n-1) and dv/dx = a/n

So dA/dx = nv^(n-1) x a/n = av^(n-1) = a(1-a/n+xa/n)^(n-1)

Answer 9

A(x)=(f(x))^n
A`(x)=nf(x)^(n-1) x f `(x)
A`(x)=nf(x)^(n-1) x a/n
A`(x) = n(1-a/n+ax/n)^(n-1) x a/n
A `(x) = a(1-a/n+ax/n)^(n-1)