Using the definition of a derivative find:

Question

y=x³-5x+1

 

(a)Find y'(1) using the denition of the derivative (i.e. the limit of a difference
quotient).
(b)Write the equation of the tangent line at x= 1

Answer 1

Part a

 

y' = [ƒ(x+h)- ƒ(x)]/h , where y=ƒ(x) and h=0

 

After finding derivative, substitute x=1

 

Part b

 

Substitute x = 1 into the original equation to get your y value. 

 

Then use the slope-intercept equation:    (y-y₁)=m(x-x₁) , 

x₁=1

y₁= 1-5+1=-5

 

Solve for y

 

%MCEPASTEBIN%

Answer 2

Jason,

 

(a) we'll write the formula for the slope between two values of x, x=1 and x = (1 + h) where h > 0:

 

slope = Δy/Δx = [y(1+h) - y(1)]/[(1+h) - 1] = [ {(1+h)³ -5(1+h) + 1} - {(1)³ -5(1) + 1}] / h

        = [ {(1 + 3h + 3h² +h³) - 4 - 5h} - {-3}] / h

        = [(1-4+3) -2h + 3h² + h³] / h

        = [ -2 + 3h + h² ]

 

Now take the limit of both sides as h approaches zero, then slope approaches the derivative y'(1) at x=1, and 3h & h2 approach zero.  Thus,

 

y'(1) = -2

 

To check this, dy/dx = y'(x) = 3x² - 5 , and when x=1, the y'(1) = -2  so checks.

 

(b)  The tangent line is the slope at a single point (whereas in geometry the slope is between 2 distinct points). And we already found it in the check above.

 

tangent = dy/dx = y'(x) = 3x2 - 5

        

Answer 3

lim

h->0     f(x+h)-f(x)

            -------------

                 h

 

That is the definition of a derivative.  

 

All you have to do is plug x+h into wherever there is an x

 

 

so you will get (x+h)³-5(x+h)+1 - (x³ -5x +1)  all divided by h with the lim as h approaches 0

 

You wil then expand out

 

x³ + 3x²h + 3xh²+h³-5x-5h+1 -x³+5x-1

 

x³ and -x³ cancel each other out  +1 and -1 cancel each other out -5x and +5x cancel each other out so you will get an equation with "h"'s in it.  

 

3x²h +3xh²+h³-5h

 

That is the top.  Now we will put this back into the limit equation

 

 

lim              3x²h+3xh²+h³-5h

h->0           ----------------------

                             h

 

You can factor out one h from the top to get  h that will cancel out with the h from the bottom to get 

 

3x² + 3xh +h²-5

 

You set the limit as h approaches 0 of this

 

and plug in 0 wherever there is an h to get

 

3x² + 3x(0) + (0)²-5

 

Your derivative to this is 3x²-5

 

 

It's not so easy to type this out, but I hope it helps!!