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QUESTION: I am really having trouble with differentiate.
Here are some questions I am having trouble with. Can you please explain to me how to do this. Thank you in advance, I really appreciate this.

Questions which i'm particularly having trouble with:
Differentiate
1/2x^3,    xsquare root x,     small 3 square root x^2

small three square root (7-3x)^2


ANSWER: On 1/2x^3, is that [A] 1/(2x^3) or [B] (1/2) of x^3, which should be put as 2x^3/3?

For [A], if f(x) = 1/(2x^3), that's the same as saying f(x) = (x^-3)/2.
From that, f'(x) = -3(x^-4)/2 = -3/(2x^4).

For [B], the derivatitve is (1/2) of 3x^2.  That's 3(x^2)/2.

For x√x, √x is the same as x^(1/2).  When x^1 is multiplied by x^(1/2),
the answer for that part is x^(3/2).  If f(x) = x^(3/2), then f'(x) = (3/2)x^(3/2 - 1).
Now 3/2 - 1 is that same as 3/2 - 2/2 = 1/2.  So the answer would be 3x^0.5/2 (that is
3(x^0/5)/2.

A small three in front of a squareroot means the cube root of x^2.
The cube root of x is the same as x^(1/3).
This means that the cube root of x^2 is the same as x^(2/3)/
If f(x) = x^(2/3), f'(x) = 2x^(-1/3)/3 = 2/(3x^(1/3)), or 2/(3(x^(1/3))).


In math, note that the nth root of x is the same as x^(1/n).
Ex: the 4th root of x = x^(1/4).

Also note that x^-n is the same as 1/x^n.
Ex: When 3 is raised to the -5 power, that is the same as 1/3^5, or 1/243.


---------- FOLLOW-UP ----------

QUESTION: Thanks that really helped.

Im having trouble with one more thing. Please explain as you did before, that was really good.

Differentiate:
x/(4x+5)^7


the answer
is -28x/(4x+5)^8         +       1/(4x+5)^7


What i dont get is my text book has gone one step further and gotten an answer of (-24x+5)/(4x+5)^8

how and why did it get that answer?

thank you

Answer
Multiply the 1/(4x+5)^7 by (4x+5)/(4x+5).  That gives (4x+5)/(4x+5)^8.

Now that this has a common denominator with -28x/(4x+5)^8, they can be added together.

That gives (-28x + 4x+5)/(4x+5)^8.

Since -28x + 4x = -24, the fraction can be rewritten as (-24x + 5)/(4x+5)^8.  

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