Thread: Derivs heeelp

f(x)=[(x+2)/(x+1)](2x-7)

I keep getting [-2x^2+7x-3] / (x+1)
but that is not one of the answer choices.

Also, (x-2)/sqrt(x^2-1)

Find an equation for the tan line to the graph f at the point P(pi/4, f(pi/4)) when f(x)=9tan x

Let f(x) = (4x-1)^6(3x+3)^2 Find the x-intercept of the tan line to the graph of f at the pt where the graph of f crosses the y-axis

Hi

Originally Posted by bballa273

f(x)=[(x+2)/(x+1)](2x-7)

I keep getting [-2x^2+7x-3] / (x+1)
but that is not one of the answer choices.

this is indeed really hard

First: Quotient rule



----
(2x-7)' = 2 =: u'

Product rule (u*v)' = u'*v + v'*u

and so on...

But it is much easier, if you multiply (x+2)(2x-7) first. Then you only must use the quotient rule

Originally Posted by bballa273

Also, (x-2)/sqrt(x^2-1)

quotient rule:

here: u= x-2 => u'=1





...

Originally Posted by bballa273

Find an equation for the tan line to the graph f at the point P(pi/4, f(pi/4)) when f(x)=9tan x

Let f(x) = (4x-1)^6(3x+3)^2 Find the x-intercept of the tan line to the graph of f at the pt where the graph of f crosses the y-axis

I dont know. Maybe someone else does

Rapha

Originally Posted by bballa273

[snip]
Find an equation for the tan line to the graph f at the point P(pi/4, f(pi/4)) when f(x)=9tan x

[snip]

You know two things about the line:

1. Point on line is (why?)

2. Gradient of line is (why?)

So you should be able to get the equation of the line.

Originally Posted by bballa273

[snip]
Let f(x) = (4x-1)^6(3x+3)^2 Find the x-intercept of the tan line to the graph of f at the pt where the graph of f crosses the y-axis

1. Find the y-intercept of y = f(x). This gives you a known point on the tangent line.

2. Evaluate the derivative of f(x) at the y-intercept. This gives you the gradient of the tangent line.

3. Use 1. and 2. to get the equation of the tangent line.

4. Use the equation in 3. to get the x-intercept of the tangent.