[SOLVED] Gradient of a Tangent.

**Awsom Guy** · October 21st 2009, 12:33 AM

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.

**mathaddict** · October 21st 2009, 12:38 AM

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

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

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

$<br /> =\frac{1}{6}<br />$

**Awsom Guy** · October 21st 2009, 12:44 AM

where's the 2 from.
Please explain.
Thanks

**Awsom Guy** · October 21st 2009, 12:59 AM

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

**calum** · October 21st 2009, 01:33 AM

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: $\frac{d}{dx} ax^n = nax^{n-1}$ where a and n are constants.

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

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

**Awsom Guy** · October 21st 2009, 01:37 AM

thanks I get that.

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[SOLVED] Gradient of a Tangent.

?

hold on..

thanks

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