Dervitative Question? Help?!?

Question

Find f '(a).

f(x) = sqrt(8x+1)

Do you use the def. of a derivate to solve? I tried that but my answer wasnt right...

Why did Kemmy put .8 in the exponent..where did that come from?

Answer 1

lim [√(8x + 8h + 1) - √(8x + 1)] / h ... as h approaches 0

Note: multiply both numerator and denominator by:
[√(8x + 8h + 1) + √(8x + 1)]

the result is:
= lim [8x + 8h + 1 - 8x -1] / h[√(8x + 8h + 1) + √(8x + 1)]
= lim 8h / h[√(8x + 8h + 1) + √(8x + 1)]
= lim 8/ [√(8x + 8h + 1) + √(8x + 1)]
= 8/ 2√(8x + 1)

= 4 /√(8x + 1).


d:

Edit: if you asked why kemmy did it... Chain Rule.

Answer 2

f(x) = sqrt(8x + 1)

If you haven't learned the various differentiation rules yet, then you would have to solve this using the definition of the derivative. Let's demonstrate that.

f'(x) =

lim [ f(x + h) - f(x) ] / h
h -> 0

Plugging in our function,

lim [ sqrt(8(x + h) + 1) - sqrt(8x + 1) ]/h
h -> 0

Multiply top and bottom by the top's conjugate. This results in a difference of squares on the top, thus eliminating the radicals.

lim [ 8(x + h) + 1 - (8x + 1) ] / [ h (sqrt(8(x + h) + 1) + sqrt(8x + 1) ]
h -> 0

Expand and simplify the numerator,

lim [ 8x + 8h + 1 - 8x - 1 ] / [ h (sqrt(8(x + h) + 1) + sqrt(8x + 1) ]
h -> 0

Notice how the 8x and -8x cancel each other out, as well as the +1 and -1.

lim [ 8h ] / [ h (sqrt(8(x + h) + 1) + sqrt(8x + 1) ]
h -> 0

Now, notice how we have h as a factor on the top and bottom. Cancelling them out, we get

lim [ 8 ] / [ (sqrt(8(x + h) + 1) + sqrt(8x + 1) ]
h -> 0

And now we can safely plug in h = 0.

[ 8 ] / [ (sqrt(8(x + 0) + 1) + sqrt(8x + 1) ]

[ 8 ] / [ (sqrt(8x + 1) + sqrt(8x + 1) ]

[ 8 ] / [ 2sqrt(8x + 1) ]

Reducing the 8 with the 2, we get

4/sqrt(8x + 1)

Answer 3

Hi,

Well, it all depends, what do the directions order you to do? If they state to use the definition, use the definition of differentiation which makes you take the limit as h goes to zero of f(x+h) - f(x) / h. However, if your teacher allows you to use the Power Rule for Differentiation, you can simply use the rule:

First, re-write the problem as:

f(x) = (8x+1)^(1/2)

Now, remember that you bring the power out in front and subtract the original power by 1. Then multiply by the derivative of the function contained inside.

Therefore, your work would look like the following:

f ' (x) = (1/2)(8)(8x+1)^(-1/2)

f ' (x) = 4 / sqr (8x+1) <=== FINAL ANSWER

NOTE: You get the final answer by simplifying (0.5)(8) to get 4 and by noticing that a negative power can be re-written in the denominator of the fraction.

I hope that helps you out! Please let me know if you have any other questions!

Sincerely,

Andrew

Answer 4

d/dx u^n = n * u^(n - 1) d/dx u


f'(x) = sqrt (8x + 1) = (8x + 1)^(1/2)

use the rule above
f'(x) = 1/2 * (8x + 1)^(1 - 1/2) d/dx (8x + 1)

f'(x) = 1/2 * ( d/dx 8x + d/dx 1) / (8x + 1)^(1/2)

f'(x) = 8 / ( 2 (8x + 1)^(1/2) )

f'(x) = 4 / (8x + 1)^(1/2)

Answer 5

y=[8x+1]^1/2

let u=8x+1, so y=u^1/2, now:

dy/dx=dy/du*du/dx, so, since

dy/du=1/2u^-1/2, and du/dx=8, then,

dy/dx=[1/2u^-1/2]*8

=1/2[8x+1]^-1/2*8

=4[8x+1]^-1/2, so, fdashed[a], the value of dy/dx when

x=a is 4[8a+1]^-1/2

Answer 6

f'(a)=1/2*(8x+1)^(-1/2)*8

Used the chain rule for derivatives.

Answer 7

f(x) = sqrt(8x+1)
f'(x) = 8/[2sqrt(8x+1)]
f'(a) = 4/sqrt(8a+1)

Answer 8

f(x) = (8x+1)^(1/2)
f'(x) = (1/2)(8x+1)^(-1/2) . 8
f'(x) = 4(8x+1)^(-1/2)
f'(a) = 4(8a+1)^(-1/2)
f'(a) = 4/[sqrt(8a+1)]

Answer 9

I presume you mean f'(x)? ; )

Think about it as f(x) = (8x+1) ^ 0.5 and see if that helps... happy to give more help if still required...

Answer 10

Answer:
f(x)=√(8x+1)
f'(x)=d/dx(√(8x+1))
Let y = 8x+1
d/dx(√(8x+1))
=d/dy(√y)*dy/dx(y)
=-1/2√y*dy/dx(8x+1)
=-1/2√(8x+1)*8
=-4/√(8x+1)