Why is this another way of writing the formula for the tangent line?

*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.

Re: Why is this another way of writing the formula for the tangent line?

Quote:

Originally Posted by

**Paze** *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 this is 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.

I hope you can see that the right hand side is actually an equation of a line, and yes it is an equation for the tangent line. What is your usual definition for the tangent line?

Re: Why is this another way of writing the formula for the tangent line?

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

Re: Why is this another way of writing the formula for the tangent line?

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Originally Posted by

**Gusbob** I hope you can see that the right hand side is actually an equation of a line, and yes it is an equation for the tangent line. What is your usual definition for the tangent line?

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.

Re: Why is this another way of writing the formula for the tangent line?

Quote:

Originally Posted by

**Paze** 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 is the limit as delta x approaches 0 of (f(x+deltax)-f(x))/deltax if that helps you understand my stubborn mindset.

If I recall correctly, this is how high schoolers are taught: to find a tangent line at the point $\displaystyle (a,f(a))$, you take $\displaystyle L=\displaystyle{\lim_{\Delta x\to 0}} \frac{f(a+\Delta x)-f(a)}{\Delta x}$. This gives you the slope of the tangent line. Since the tangent line passes through the point $\displaystyle (a,f(a))$, the point-slope formula gives $\displaystyle y-f(a)=L(x-a)$, which can be rearranged as $\displaystyle y=f(a)+L(x-a)$. Here's the kicker: $\displaystyle f'(a)=L=\displaystyle{\lim_{\Delta x\to 0}} \frac{f(a+\Delta x)-f(a)}{\Delta x}$ __by definition__, so you actually do get $\displaystyle y=f(a)+f'(a)(x-a)$.

Re: Why is this another way of writing the formula for the tangent line?

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.

Re: Why is this another way of writing the formula for the tangent line?

Quote:

Originally Posted by

**Gusbob** If I recall correctly, this is how high schoolers are taught: to find a tangent line at the point $\displaystyle (a,f(a))$, you take $\displaystyle L=\displaystyle{\lim_{\Delta x\to 0}} \frac{f(a+\Delta x)-f(a)}{\Delta x}$. This gives you the slope of the tangent line. Since the tangent line passes through the point $\displaystyle (a,f(a))$, the point-slope formula gives $\displaystyle y-f(a)=L(x-a)$, which can be rearranged as $\displaystyle y=f(a)+L(x-a)$. Here's the kicker: $\displaystyle f'(a)=L=\displaystyle{\lim_{\Delta x\to 0}} \frac{f(a+\Delta x)-f(a)}{\Delta x}$ __by definition__, so you actually do get $\displaystyle y=f(a)+f'(a)(x-a)$.

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.

Re: Why is this another way of writing the formula for the tangent line?

I'll demonstrate that the two are the same thing:

I'm going to set $\displaystyle L=f'(a)$ so the following is easier to read. When we take the derivative of $\displaystyle f(x)$ at $\displaystyle a$, we're finding the slope of the tangent line at $\displaystyle a$, which we have established to be $\displaystyle L=f'(a)$. Using your method, you have $\displaystyle y=Lx+b$ where $\displaystyle b$ is the $\displaystyle y$ intercept. To find $\displaystyle b$, we substitute the only point we know which satisfies this equation: $\displaystyle (a,f(a))$. This gives $\displaystyle f(a)=La+b \implies b=f(a)-La$. Therefore the tangent line is $\displaystyle y=Lx+b=Lx+f(a)-L(a)=f(a)+L(x-a)$.

Re: Why is this another way of writing the formula for the tangent line?

Quote:

Originally Posted by

**Paze** *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.

It would have helped if, when you first posted this you told us what you thought the "only definition" for a tangent line **was**! What you give here is, I think, what most people first see as "the tangent line to f(x) at (x0, f(x0))".