kalisto_911

find y''' if y=(2x+3)^1/2 (or the square root of 2x+3)

the answer in the book is 3(2x+3)^-5/2 but I can't seem to arrive at this, my number where the first 3 is located is different.

Calc midterm tomorrow (and then final tues.!) and I hate learning stuff like this all in one day

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Rekna

f'(x)=1/(2*x+3)^(1/2)
f''(x)=-1/(2*x+3)^(3/2)
f''(x)=3/(2*x+3)^5/2

make sure each derivitive you are taking is correct.

kalisto_911

can you explain a bit more?

if the derivative of x^n = nx^n-1 then wouldn't f'(x)=1/2(2x+3)^-1/2 ? I guess having the power outside the parenthesis is confusing me.

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I know not how I may seem to others, but to myself I am but a small child wandering upon the vast shores of knowledge, every now and then finding a small bright pebble to content myself with. [Plato]

sandinista

That's a chain rule problem. If f(x) = (x^n)^n then f'(x) = (n(x^n)^n-1)*nx^n-1 - you get it? for that problem,

y=(2x+3)^1/2
y'= (1/2(2x+3)^-1/2)*2
y''=2((-1/4(2x+3)^-2)*2) and so on.

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kalisto_911

Ok I think I got this problem licked, thanks for the help and wish me luck tomorrow

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I know not how I may seem to others, but to myself I am but a small child wandering upon the vast shores of knowledge, every now and then finding a small bright pebble to content myself with. [Plato]

Slavakion

Is there a shortcut to finding second/third/... derivatives? Not that it's difficult (yet), but it gets tedious. Especially when you get the evil ones with multiple chain rules.

shred_head

I don't think that there is, I think you just have to suck it up and punch it out. However, if you're dealing with trig or exponential functions you can occasionally rewrite the derivates in terms of the original function. Such as

f(x) = sin x
f'(x) = cos x
f''(x) = - sin x = - f(x)
f'''(x) = - f'(x) = - cos x
and so on...

But for the most part I think you just have to do it the long way.

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phukraut

Well, you can memorize certain formulas. For example, in the case of the nth derivative of x^k, where n is a nonnegative integer, we have

D^n x^k = (k)_n x^{k-n}, where

(k)_n is the falling factorial of k to n, namely,

(k)_n = k (k-1) (k-2) ... (k-n+1) if n>0;
(k)_0 = 1.

Another useful formula is Leibniz' Rule, which is a generalization of the product rule, and looks like this:

D^n (f(x)g(x)) = \sum_{i=0}^n \binom{n}{i} f^(i)(x) g^(n-i)(x),

where \binom{n}{i} is the binomial coefficient, and f^(i) is the ith derivative of f. Note its relation to the binomial theorem, making it easy to remember. Graphically,



There's also a generalization of the chain rule, but I can't remember it and it's sort of ugly anyway.

sandinista

there's also functions like f(x) = x/(y)^z, f'(x)=x*-z/(y)^z+1, but as a rule, there's no consistent way to take higher derivatives.

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"Nature herself makes the wise man rich."
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