# Math Help - Calculus HW due in 3 hours (4 Problems I cant get)

1. ## Calculus HW due in 3 hours (4 Problems I cant get)

Well I've been trying these problems for awhile, I'll type the problem out as well as what I have gotten as the answer and how I got there. Thanks for any help...my prof likes to throw things at us that is very challengin on HW..
PROBLEM 1
f(x)=1/(x-2x^2)^(9/8)
Find the derivative of f(x).

For starters i moved the binomial to the numerator and switched the power to negative giving me f(x)=(x-2x^2)^(-9/8) . After that I simply applied the chain rule giving me (-9/8)(x-2x^2)^-17/8(-4x) . It says this answer is incorrect...

PROBLEM 2
f(x)=sqrt(3+sqrt4x) The 4x has two sqrt over it. I broke this problem down to read f(x)=(3^1/2+4x^1/2)^1/2 . I then used the chain rule which gave me (1/2)(3^1/2+4x^1/2)(2x^-1/2) . It says this answer is incorrect...

I'll post the other two up after I get some help on these because I havent even been able to start them. They are more difficult then these two. Thanks for the help.

You made some small errors . . .

PROBLEM 1: .f(x) .= .1/(x - 2x²2)^{9/8}
Find the derivative of f(x).

For starters i moved the binomial to the numerator and switched the power
to negative, giving me: .f(x) .= .(x - 2x²)^{-9/8}

After that I simply applied the chain rule giving me: .(-9/8)(x - 2x²)^{-17/8}(-4x) . no
The derivative of the "inside" is: .1 - 4x

PROBLEM 2: .f(x) .= .sqrt[3 + sqrt(4x)]

I broke this problem down to: .f(x) .= .(3 + 4x^½)^½ . no
The inner square root is on the entire 4x.

. . f(x) .= .[3 + (4x)^½]^½ .= .(3 + 2x^½)^½

Now give it a try . . .

3. Originally Posted by lmao
Well I've been trying these problems for awhile, I'll type the problem out as well as what I have gotten as the answer and how I got there. Thanks for any help...my prof likes to throw things at us that is very challengin on HW..
PROBLEM 1
f(x)=1/(x-2x^2)^(9/8)
Find the derivative of f(x).

For starters i moved the binomial to the numerator and switched the power to negative giving me f(x)=(x-2x^2)^(-9/8) . After that I simply applied the chain rule giving me (-9/8)(x-2x^2)^-17/8(-4x) . It says this answer is incorrect...

PROBLEM 2
f(x)=sqrt(3+sqrt4x) The 4x has two sqrt over it. I broke this problem down to read f(x)=(3^1/2+4x^1/2)^1/2 . I then used the chain rule which gave me (1/2)(3^1/2+4x^1/2)(2x^-1/2) . It says this answer is incorrect...

I'll post the other two up after I get some help on these because I havent even been able to start them. They are more difficult then these two. Thanks for the help.

Q1 f(x)=(x-2x^2)^(-9/8)

Let u = (x-2x^2)
u' = 1-4x
f(x) = u^(-9/8)
df(x)/du = (-9/8)u^(-17/8)

Then df(x)/dx = (1-4x)(-9/8)u^(-17/8)
= (1-4x)(-9/8)[(x-2x^2)^(-17/8)]

Q2 f(x)=sqrt(3+sqrt4x) = [ 3+((4x)^1/2) ] ^ (1/2)

Let u = 3+((4x)^1/2)
u' = (2x)^(-1/2)
f(x) = u ^ (1/2)
df(x)/du = (1/2)u^(-1/2)

Then df(x)/dx = (2x)^(-1/2) . (1/2)u(-1/2)
= (2x)^(-1/2) . (1/2)[3+((4x)^1/2)]^(-1/2)

That's very ugly, but I'm sure you can simplify it in some way.

4. Originally Posted by lmao
Well I've been trying these problems for awhile, I'll type the problem out as well as what I have gotten as the answer and how I got there. Thanks for any help...my prof likes to throw things at us that is very challengin on HW..
PROBLEM 1
f(x)=1/(x-2x^2)^(9/8)
Find the derivative of f(x).

For starters i moved the binomial to the numerator and switched the power to negative giving me f(x)=(x-2x^2)^(-9/8) . After that I simply applied the chain rule giving me (-9/8)(x-2x^2)^-17/8(-4x) . It says this answer is incorrect...

PROBLEM 2
f(x)=sqrt(3+sqrt4x) The 4x has two sqrt over it. I broke this problem down to read f(x)=(3^1/2+4x^1/2)^1/2 . I then used the chain rule which gave me (1/2)(3^1/2+4x^1/2)(2x^-1/2) . It says this answer is incorrect...

I'll post the other two up after I get some help on these because I havent even been able to start them. They are more difficult then these two. Thanks for the help.

For Problem 1.
The derivative of (x -2x^2) is (1 -4x). Not (-4x) only.

For Problem #2.
Your f(x) = [(3^(1/2) +4x^(1/2)]^(1/2) is really incorrect.
What is the name of that property? Sorry, I am not well-versed on definitions, but you cannot "spread" the sqrt to each term in the radicand that are in summation.
sqrt(9 +16) is not sqrt(9) + sqrt(16). -----that gives 3 +4 = 7.
sqrt(9 +16) is sqrt(25). -----that gives 5, which is correct.

f(x) = sqrt[3 +sqrt(4x)]
So,
f(x) = [3 +(4x)^(1/2)]^(1/2)
then,
f'(x) = (1/2)[3 +(4x)^(1/2)]^(-1/2) *(1/2)[(4x)^(-1/2) *4]
f'(x) = (1/2)(1/2)(4)[3 +(4x)^(1/2)]^(-1/2) *[(4x)^(-1/2)]
f'(x) = [3 +(4x)^(1/2)]^(-1/2) *[(4x)^(-1/2)]
Since sqrt(a) *sqrt(b) = sqrt(a*b), then,
f'(x) = [{3 +(4x)^(1/2)}*(4x)]^(-1/2)

I intentionally did not reduce the sqrt(4x) in 2sqrt(x) yet so that you won't be more confused.
Here, sqrt(4x) = sqrt(4) *sqrt(x). You can "spread" here the sqrt to each term in the radicand because those terms are in multiplication.
sqrt(9 +16) is not the same as sqrt(9*16)
sqrt(9*16) = sqrt(9) *sqrt(16) = 3*4 = 12.
sqrt(9*16) = sqrt(144) = 12.

So,
f'(x) = [{3 +2sqrt(x)}*(2sqrt(x))]^(-1/2)
f'(x) = [3*2sqrt(x) +[2sqrt(x)]*2sqrt(x)]^(-1/2)
f'(x) = [6sqrt(x) +4x]^(-1/2)
f'(x) = 1 / sqrt[6sqrt(x) +4x] ---------------answer.

5. Originally Posted by lmao
...

PROBLEM 2
f(x)=sqrt(3+sqrt4x) The 4x has two sqrt over it. I broke this problem down to read f(x)=(3^1/2+4x^1/2)^1/2 . I then used the chain rule which gave me (1/2)(3^1/2+4x^1/2)(2x^-1/2) . It says this answer is incorrect...
...

I used the result Soroban has given. Then use the chain rule very carefully:

6. Originally Posted by lmao
Well I've been trying these problems for awhile, I'll type the problem out as well as what I have gotten as the answer and how I got there. Thanks for any help...my prof likes to throw things at us that is very challengin on HW..
PROBLEM 1
f(x)=1/(x-2x^2)^(9/8)
Find the derivative of f(x).

For starters i moved the binomial to the numerator and switched the power to negative giving me f(x)=(x-2x^2)^(-9/8) . After that I simply applied the chain rule giving me (-9/8)(x-2x^2)^-17/8(-4x) . It says this answer is incorrect...

PROBLEM 2
f(x)=sqrt(3+sqrt4x) The 4x has two sqrt over it. I broke this problem down to read f(x)=(3^1/2+4x^1/2)^1/2 . I then used the chain rule which gave me (1/2)(3^1/2+4x^1/2)(2x^-1/2) . It says this answer is incorrect...

I'll post the other two up after I get some help on these because I havent even been able to start them. They are more difficult then these two. Thanks for the help.