We’re starting a new unit: applications of diﬀerentiation.
We’re going to do two applicationstoday. The ﬁrstislinear approximations,
which are encompassed by the single formula:
f(x) ≈ f(x0)+f'(x0)(x−x0),
but it will take at least half an hour to explain how this formula is used.
The teacher explains that the right hand side is the equation for the tangent line? Or am I misunderstanding?
If I am not, how come this is the equation for the tangent line? I thought there was only one definition for that.
Paze
Study little the tangent lines
the equation of the tangent TO A CURVE AT THE POINT (X0,Y0) is y-y0=(gradient)(x-x0)
the value of the derivative f'(x0) at the point (x0,y0) is the gradient .therefore the formula is :
y=y+f"(x0)(x-x0) it is simple.....
MINOAS
Thank you guys, but I am being very obtuse unfortunately. I have proven to myself in geogebra that you are in fact right. However, I am having problems visualizing this.
My usual definition for the tangent line of a function at any given point 'x' is the limit as delta x approaches 0 of (f(x+deltax)-f(x))/deltax if that helps you understand my stubborn mindset.
For general derivatives I simply use the power rule, chain rule and all that which I have already proven before...This one must have slipped me.
If I recall correctly, this is how high schoolers are taught: to find a tangent line at the point , you take . This gives you the slope of the tangent line. Since the tangent line passes through the point , the point-slope formula gives , which can be rearranged as . Here's the kicker: by definition, so you actually do get .
I am understanding this better now. I am finding the EQUATION of the tangent line, duh. Not the slope.
However I am still trying to understand how this formula does that (I need to prove it). What I usually do is just take the derivative at a point and then do the good ol' y=kx+m routine to find the equation.
Yes, thank you.
The only thing I now need is to understand this 'point-slope' formula. I usually chug through these things with deductive reasoning so I am not used to seeing formulas that do things for me :/
I found a proof for the point-slope formula. Thank you very much.
I'll demonstrate that the two are the same thing:
I'm going to set so the following is easier to read. When we take the derivative of at , we're finding the slope of the tangent line at , which we have established to be . Using your method, you have where is the intercept. To find , we substitute the only point we know which satisfies this equation: . This gives . Therefore the tangent line is .