# Understanding the derivative geometrically

I always seen the derivative of a function $y=f(x)$,$\frac{dy}{dx}$ at $x_1$ as the slope of the line tangent to the curve $y=f(x)$ drawn at $y=f(x_1)$.But I often fail to appreciate this when $\frac{dy}{dx}=0$ at some point $x_1$ .

Can anyone please tell me the geometrical significance of $\frac{dy}{dx}=0$

or

draw an analogy which would apply to the above-mentioned case?

(In fact, analytically, what is a tangent to a curve?)

Sorry for so many weird questions.I hope I am not being too incoherent.

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When the derivative of a function $f$ is $0$ at say $(x_0,f(x_0))$, then that means that you have a horizontal like tangent to this point $(x_0,f(x_0))$ of the function.

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I would comment, but lack reputation. http://upload.wikimedia.org/wikipedia/commons/7/7a/Graph_of_sliding_derivative_line.gif May help intuitively: $\frac{dy}{dx}=0$ corresponds to when the tangent is black in the graphic (i.e. horizontal).

As to your first question, mathematically the tangent to $f(x)$ at $a$ is $y=f(a)+f'(a)(x-a)$, which pops out naturally from the definition of the derivative, (i.e. $f'(a)=lim, x \rightarrow a (\frac{f(x)-f(a)}{x-a})$, as a tangent is (I think) the linear approximation to a function from a point (being more of an accurate approximation nearer the point).

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Just to add more informally, derivative as well as slope shows how much $y$ changes for a "small" change in $x$. For a straight-line, this ratio doesnt change even if change in $x$ were not small, hence slope or derivative is constant. For differentiable curves in general, this ratio would be unique if you could zoom in to the curve as much as you want. In other words, differentiable functions would have any point $a$, another points (infinitly) close to it on left,$a^-$ and on right,$a^+$ such that all three of them, $a^-, a, a^+$ lie on straight line which is tangent to the curve at this point $a$.

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