simple derivatives?

**jayjay60** · September 26th 2009, 09:25 AM

i absolutely LOVE derivatives. i CANNOT figure this one out. i have major trouble with square roots. >,<!!

f(x)= 1/√(x+2),, so find f'(a)
yeah. i know. simple right? i just can't figure out. i know its ^1/2 for the sqrt, but blehhhhh! i've tried this one a million times!

i know the answer is -1/2(a+2)^3/2, back of the book. but erggg >.<

well, while we're at it-->this one too?! just a simple derivation. it's just, "whaaa? why is THAT not right?" lol

**TheEmptySet** · September 26th 2009, 09:32 AM

Originally Posted by jayjay60

i absolutely LOVE derivatives. i CANNOT figure this one out. i have major trouble with square roots. >,<!!

f(x)= 1/√(x+2),, so find f'(a)
yeah. i know. simple right? i just can't figure out. i know its ^1/2 for the sqrt, but blehhhhh! i've tried this one a million times!

i know the answer is -1/2(a+2)^3/2, back of the book. but erggg >.<

well, while we're at it-->this one too?! just a simple derivation. it's just, "whaaa? why is THAT not right?" lol

$f(x)=\frac{1}{\sqrt{x+2}}=(x+2)^{-1/2}$

$f'(x)=-\frac{1}{2}(x+2)^{-3/2}=-\frac{1}{2(x+2)^{3/2}}$

For the next one we need either the chain rule or a little algebra

Here is the algebra first.

$F(x)=\left( \frac{1}{2}x\right)^4=\frac{1}{2^4}x^4=\frac{1}{16 }x^4$

$F'(x)=\frac{1}{4}x^3$

Now by the chain rule we get

$F(x)=\left( \frac{1}{2}x\right)^4$

$F(x)=4\left( \frac{1}{2}x\right)^3\cdot \left( \frac{1}{2}\right)=\frac{1}{4}x^3$

**jayjay60** · September 26th 2009, 09:57 AM

thank you so much! i was getting 2x^3...i was putting 2x instead of 1/4...silly algebraic mistakes. they get me every time!

i'm very grateful indeed!

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