[SOLVED] derivatives

**Corko93** · February 4th 2009, 06:07 AM

So the question is.. Find the gradient of the following curves at the given point.

f(x) = 5 / (root x) at the point where x=9

I know that Root x = x^1/2

But the numerator being 5 has confused me... Where would I go from here?
Thankyou for any help.

**Moo** · February 4th 2009, 06:10 AM

Hello,

Originally Posted by Corko93

So the question is.. Find the gradient of the following curves at the given point.

f(x) = 5 / (root x) at the point where x=9

I know that Root x = x^1/2

But the numerator being 5 has confused me... Where would I go from here?
Thankyou for any help.

You have to find $f'(9)$

The 5 is not disturbing. Because the derivative of $af(x)$ , where a is a constant is $af'(x)$

So $\left[\frac{5}{\sqrt{x}}\right]'=5 \cdot \left[\frac{1}{\sqrt{x}}\right]'$

Can you calculate this derivative ?

**Corko93** · February 4th 2009, 07:31 AM

So that would be 5x^1/2

which differentates to 2.5x^-1/2 ?
I think...

P.S - I'm sorry I haven't learned the codes for correct displaying of the expression

**Moo** · February 4th 2009, 08:49 AM

Originally Posted by Corko93

So that would be 5x^1/2

which differentates to 2.5x^-1/2 ?
I think...

P.S - I'm sorry I haven't learned the codes for correct displaying of the expression

Nope, it's $5x^{-1/2}$ (click on the picture to get the code)

So it differentiates to $5 \cdot \left(-\frac 12\right) x^{-1/2-1}=-2.5 x^{-3/2}$

**Corko93** · February 4th 2009, 10:14 AM

So when I substitute the x value back in to work out my gradient

$ -2.5 (9)^{-3/2} = -0.092592596 $

Either I'm going wrong somewhere or I'm supposed to represent that as a surd?

**Moo** · February 5th 2009, 09:58 AM

Originally Posted by Corko93

So when I substitute the x value back in to work out my gradient

$ -2.5 (9)^{-3/2} = -0.092592596 $

Either I'm going wrong somewhere or I'm supposed to represent that as a surd?

$9^{-3/2}=\frac{1}{9^{3/2}}=\frac{1}{(9^{1/2})^3}=\frac{1}{3^3}=\frac{1}{27}$

**Corko93** · February 6th 2009, 09:49 AM

Well that would explain alot!
Thankyou Moo!

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