Derivative of f(x)=sqrt(7+sqrtx^3)

[Buzzlastyear]

Homework Statement

Determine the derivative of f(x)=√(7+√x^3)

Homework Equations

Chain rule: f(x)=fog(x) f'(x)=f'(g(x)) x g'(x)

The Attempt at a Solution

f(x)=√(7+x^(3/2))

f'(x)=((½)x^(-½))(7+x^(3/2))((3/2)x^(½))
=(3.5x^(-½)+½x)((3/2)x^(½))
=5.25x^0+¾x^½
=5.25+¾x^½

That's as far as i got and i feel like it is completely wrong. Any guidance and help is very much appreciated! Thanks in advance.

 

[Infinitum]

f'(x)=((½)x^(-½))(7+x^(3/2))((3/2)x^(½))

Here lies the mistake, chain rule is f(x)=fog(x) f'(x)=f'(g(x)) x g'(x)

You can take two functions in the given problem, $f(x) = \sqrt{x}$ and $g(x) = 7+\sqrt{x^3}$

If you differentiate $\sqrt{7+\sqrt{x^{3}}}$ following the chain rule, you would get the first term f'(g(x)) as,

$\frac{1}{2\sqrt{g(x)}}$

See your mistake?

 

[Buzzlastyear]

Ah yes, thank you!

 

[Buzzlastyear]

So would it now be f'(x)=(1/(2√7+√x^3))((3/2)x^(1/2))?

 

[Mark44]

Chain rule: f(x)=fog(x) f'(x)=f'(g(x)) x g'(x)

This is a confusing (to be charitable) use of notation.
f(x) is not the same as (f o g)(x), and f'(x) is different from f'(g(x)).

Also, you should not use x to indicate multiplication, especially when you already have a variable named x.

A better way to write your chain rule formjula would look like this:
h(x) = f(g(x)) ==> h'(x) = f'(g(x)) $\cdot$ g'(x)

 

[Infinitum]

So would it now be f'(x)=(1/(2√7+√x^3))((3/2)x^(1/2))?

Almost...

It should be, 1/2(√7+√x^3) Parentheses are very important!