Question

Suppose that f(x+h)-f(x)/h=-2h(x+5)-h^2/h(x+h+5)^2(x+5)^2 Find the slope m of the tangent line at x=4 . m=

Answer 1

The slope of the tangent line is the derivative function with x=4. Are you familiar with derivatives or new to them? Have you learned the shorter version of the power rule for derivatives or are you still working with this?

Limit definition of a derivative:
\[ \huge \lim_{{\triangle x} \rightarrow 0} \frac{f(x+\triangle x)+f(x)}{\triangle x} \]

Answer 2

h = \( \triangle \)x for your problem, same thing, different name

Answer 3

i am still trying to figure out derivatives

Answer 4

And the second question I asked? o_O Class progress = ?

Answer 5

we have yet to get to it but i was working ahead

Answer 6

I could teach you the shortcuts, but that might mess you up on an exam depending on the teacher you have, if they want to do it the long way until THEY teach you the shortcuts

Answer 7

sounds good

Answer 8

Alrighty, let's rewrite that bad boy first, you're going to have to show parenthesis a bit more clearly because currently you have (which is not equals):
\[ \frac{f(x+h)-f(x)}{h} \neq-2h(x+5)-h^2/h(x+h+5)^2(x+5)^2 \]

Answer 9

What's on the left is golden, but whatever you have on the right I need you to rewrite please. Carefully use parenthesis to indicate order of operation the way you meant it. Cool?

Answer 10

i will note that thanks.

Answer 11

can u see the file i attached?

Answer 12

The trick is basically this, you have some function getting x input plus a little bit more and the difference of that with the little bit more

Remember this?
\[slope=\frac{rise}{run}=\frac{\triangle y}{\triangle x}\]

h = change in x = \(\triangle\)x

Answer 13

I see it. One moment...

Answer 14

Do you remember talking about average slope?

Answer 15

\[m_{average}=\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\]

Answer 16

Well what you're doing is making the differences in the two x coordinates get very very small, right down to nothing at all (limit as \(\triangle\)x goes to zero) so you get the exact INSTANTANEOUS slope, instead of average. Which is that very definition of the derivative.

With the right hand side of the equation shown in your problem, have you tried to simplify it yet? If not, you need to do that first. You're trying to make stuff cancel out, the h's specifically