1. Originally Posted by Pikeman85
Still not getting it. I'm very dense about a lot of the simpler bits. I've got most of the differentiation ones done, but this one eludes me
f(2) from left equals 4+12+13=29

f(2) from right 2a+b

f'(2) from left equals 4+6=10

f'(2) from right equals a

so $\displaystyle a=10$

and $\displaystyle 2a+b=29$

so a=10 and b=9

2. Wow, that was a lot easier than what I was trying to do. I knew it was something easy I had to do as well. The worst bit is that it's my algebra skills that have gone down over the years, over everything else.

3. Originally Posted by Pikeman85
Wow, that was a lot easier than what I was trying to do. I knew it was something easy I had to do as well. The worst bit is that it's my algebra skills that have gone down over the years, over everything else.
Well dont fret! After a course in calc your alg. skills will be given a makeover...that or you fail

4. Find the equation of the tangent line to the curve y = 2x cos x at the point (pi , -2\pi). The equation of this tangent line can be written in the form y = mx+b where

m =

b =

Now I know I have to plug this into f(x-h) - f(x)/h

or is there a different way to do it?

5. Originally Posted by Pikeman85
Find the equation of the tangent line to the curve y = 2x cos x at the point (pi , -2\pi). The equation of this tangent line can be written in the form y = mx+b where

m =

b =

Now I know I have to plug this into f(x-h) - f(x)/h

or is there a different way to do it?
Yes, the equation of a tangent line is $\displaystyle y-f(x_0)=f'(x_0)(x-x)0)$

where $\displaystyle f'(x_0)$ is the derivative evaluated at the point

6. ( f(3 x^{2}) ) = 9 x^{3}.

How do I calculate f'(x) in this case?

I swear this is the last question I ask tonight!

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