Will anyone verify this answer please? find f'(x) of f(x)=x/(x^2+7)^1/2?

Question

Using the quotient rule I came up with:

f'(x) = [-2x^2 + (x^2+7)^1/2] / (x^2+7)^3/2

It seems to me that either I screwed up or I can't figure out how to simplify this better. I appreciate your time and help.
Thanks,
Jim

Answer 1

f (x) = x / (x² + 7)^(1/2)

f `(x) is given by:-

(1)(x² + 7)^(1/2) - (x)(1/2)(x² + 7)^(-1/2)(2x)
------------------------------------------------------
(x² + 7)

(x² + 7)^(-1/2) [ (x ² + 7) - x ² ]
-------------------------------------------
x² + 7

7 / (x² + 7)^(3/2)

Answer 2

jim,

f(x)=x/(x^2+7)^1/2
we can move the denominator to the numerator
f(x)=x(x^2+7)^-1/2
now its only product rule and dont forget the chain rule
f'(x) = 1[(x^2+7)^-1/2] + x(-1/2)[(x^2+7)^-3/2](2x)
f'(x) = [(x^2+7)-1/2] + -x^2[(x^2+7)^-3/2]
f'(x) = [(x^2+7)-3/2][(x^2+7)-x^2]
f'(z) = [(x^2+7)-3/2]7

dunno.. im lost also just check

Answer 3

I never like solving these as fractions, so I'll rewrite as a product:
f(x) = x*(x^2 + 7)^(-1/2)

Differentiating:
f'(x) = (x^2 + 7)^(-1/2) + x*(-1/2)*(x^2 + 7)^(-3/2)
f'(x) = [2*(x^2 + 7)^3 - x] / [2*(x^2 + 7)^(3/2)]

Answer 4

i do no longer understand if i will clarify why, yet i will clarify why the others are fake. f(x) = sqrt(x^2) (a) x This one is obviously fake, because of the fact if we enable x = -one million, then sqrt(x^2) = sqrt( (-one million)^2 ) = sqrt(one million) = one million, for sure no longer equivalent. (b) -x. This one is fake too. If we pick x = one million, then -x = -one million, and sqrt(x^2) = sqrt(one million^2) = sqrt(one million) = one million (c) +/- x it incredibly is fake. enable x = one million. Then +/- x = { one million, -one million } yet sqrt(x^2) = sqrt(one million) = one million Demonstrating that the function sqrt is precisely a function that returns one (no longer 2) values. it is why, whilst we've an equation like x^2 = 7, we get x = +/- sqrt(7) (2 values returned) as a replace of x = sqrt(7) (one fee, the useful one, returned) So generally - in simple terms because of the fact it makes use of the sq. root, does no longer mean it incredibly is comparable to taking the sq. root of the two facets of an equation (which finally ends up in 2 ideas, in lots of cases). sqrt(9) = 3, yet x^2 = 9 x = +/- sqrt(9) x = +/- 3 sqrt refers back to the _Principle_ sq. root (that's useful). If we would have loved the destructive sq. root too, there may be a destructive sign exterior of the sq. root.

Answer 5

It should be:

7/[(x^2 + 7)^(3/2)]

I'll take you through it:::
f(x)=x/(x^2+7)^1/2
Low D high - High D low all over the square of what's below
f'(x) = (x^2+7)^(1/2) - x (1/2)(2x)(x^2 +7)^(-1/2)
-------------------------------------------------------
x^2 + 7

Simplify:
f'(x) = (x^2+7)^(1/2) - x^2(x^2 +7)^(-1/2)
-------------------------------------------------------
x^2 + 7

You can do this step here


Simplifies to:
7
---------
(x^2 + 7)^(3/2)

Answer 6

f(x)=g(x)/h(x)

quotient rule
f'(x) = [h(x)g'(x)-g(x)h'(x)]/ [h(x)]^2

the problem
f(x)=x/(x^2+7)^1/2

h(x) = x
h'(x) = 1
g(x) = (x^2+7)^1/2
to find g'(x) you need to use the chain rule
g'(x) = 1/2 (x^2+7)^1/2 (2x) = x (x^2+7)^1/2

Now you can use the
quotient rule
f'(x) = [h(x)g'(x)-g(x)h'(x)]/ [h(x)]^2
f'(x) = [(x^2+7)^1/2(1)-(x)x(x^2+7)^1/2]/ [(x^2+7)^1/2]^2
f'(x) = [(x^2+7)^1/2-x^2(x^2+7)^1/2]/ [(x^2+7)]

Answer 7

I just answered one of your other questions. You should learn the stuff and do your own homework.