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

  • #1

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.
 

Answers and Replies

  • #2
881
40
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, [itex]f(x) = \sqrt{x}[/itex] and [itex]g(x) = 7+\sqrt{x^3}[/itex]

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

[itex]\frac{1}{2\sqrt{g(x)}}[/itex]

See your mistake? :smile:
 
  • #3
Ah yes, thank you!
 
  • #4
So would it now be f'(x)=(1/(2√7+√x^3))((3/2)x^(1/2))?
 
  • #5
34,974
6,724
Buzzlastyear said:
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)
 
  • #6
881
40
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!
 

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