1. ## [SOLVED] Gradient of a Tangent.

I think this belongs here, hopefully here....

What is the gradient of the tangent to the graph of f(x)=sqrt(x) at the point where x=9.
I think the answer should be 0.
But I need the answer confirmed.
Thanks.

2. Originally Posted by Awsom Guy
I think this belongs here, hopefully here....

What is the gradient of the tangent to the graph of f(x)=sqrt(x) at the point where x=9.
I think the answer should be 0.
But I need the answer confirmed.
Thanks.
HI

nope , its not 0

$\displaystyle f'(x)=\frac{1}{2\sqrt{x}}$

$\displaystyle =\frac{1}{2\sqrt{9}}$

$\displaystyle =\frac{1}{6}$

3. ## ?

where's the 2 from.
Thanks

4. ## hold on..

i think im getting it but I cant find out how to get the inverse 1/ bit.

5. Originally Posted by Awsom Guy
i think im getting it but I cant find out how to get the inverse 1/ bit.
The power rule for differentiation is: $\displaystyle \frac{d}{dx} ax^n = nax^{n-1}$ where a and n are constants.

For this case: $\displaystyle f(x) = \sqrt{x}$ so $\displaystyle a=1$ and $\displaystyle n = \frac{1}{2}$

Thus: $\displaystyle f'(x) = \frac{1}{2} x^{\frac{-1}{2}} = \frac{1}{2\sqrt{x}}$

6. ## thanks

thanks I get that.