Question

if f(x)=sqrt(10+sqrt(8x)), f'(x)=?

Answer 1

\[\sqrt{10+\sqrt{8x}}\]

Answer 2

\[f(x)=\sqrt{10+\sqrt{8x}}\] like that?

Answer 3

yup

Answer 4

only a math teacher could like this

Answer 5

we start with the fact that
\[\frac{d}{dx}\sqrt{f}=\frac{f'}{2\sqrt{f}}\]

Answer 6

so it is going to look like
\[\frac{g'(x)}{2\sqrt{10+\sqrt{8x}}}\] where \(g(x)=10+\sqrt{8x}\)

Answer 7

alright it says find the derivative so we use the power of a function rule 2 times where(f(x))^n = n(f(x))^(n-1) * f'(x)

so we got (1/2)(10 + (8x)^(1/2)) ^(-1/2) * (1/2)(8x)^(-1/2) *8 and now we are just going to clean this mess up a get

2/[(10+(8x)^(1/2))*(8x)^(1/2)] :) if that is easy to understand

Answer 8

also we might note that \(\sqrt{8x}=2\sqrt{2x}\) so the derivative of
\[10+2\sqrt{2x}\] is
\[\frac{\sqrt{2}}{\sqrt{x}}\]

Answer 9

This sort of problem is made much easier, in my humble opinion, if we use fractional exponents instead: (10 + [8x]^0.5)^0.5. Now it's just an application of the chain rule (which may admittedly get rather confusing for this one).

But, I like the way satellite is breaking this problem down.

Answer 10

it is a good thing to memorize, since it comes up very very often, that
\[\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\] without resorting to exponential notation (although of course that works)
and therefore by the chain rule
\[\frac{d}{dx}[\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\] with no agony and no exponents

Answer 11

Good points. Being a computer science person, I suppose I try to avoid memorizing anything. :p

Answer 12

final answer (before simplifying) will be
\[\frac{\frac{\sqrt{2}}{\sqrt{x}}}{2\sqrt{10+2\sqrt{2x}}}\]

Answer 13

which becomes
\[\frac{\sqrt{2}}{\sqrt{x}}\times \frac{1}{2\sqrt{10+2\sqrt{2x}}}\]