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