# Math Help - [SOLVED] derivatives

1. ## [SOLVED] derivatives

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.

2. 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 ?

3. 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

4. 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 $5 \cdot \left(-\frac 12\right) x^{-1/2-1}=-2.5 x^{-3/2}$

5. 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?

6. 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?
$9^{-3/2}=\frac{1}{9^{3/2}}=\frac{1}{(9^{1/2})^3}=\frac{1}{3^3}=\frac{1}{27}$

7. Well that would explain alot!
Thankyou Moo!