Help with Derivative Problem

[elindow]

I have just have a question I have worked the problem and I want to know if I am right or not.

I have to find the first and second derivative and this is the what I got:

f(x)=(x²-3x-5)^1/3

this is what I got

f'(x)=(2x-3)/(x²-3x-5)^(2/3)

f"(x)=2-(2(2x-3))²/3(x²-3x-5)^1/3

is this right?

Thanks
Erik

 

[Daniel_Feldman]

I have just have a question I have worked the problem and I want to know if I am right or not.

I have to find the first and second derivative and this is the what I got:

f(x)=(x²-3x-5)^1/3

this is what I got

f'(x)=(2x-3)/(x²-3x-5)^(2/3)

f"(x)=2-(2(2x-3))²/3(x²-3x-5)^1/3

is this right?

Thanks
Erik

Hey Erik.

Careful now! You seem to know what you're doing, but practice the power/chain rules a bit more.

f(x)=(x^2-3x-5)^(1/3)

Remember the power rule: If f(x)=x^n, then f'(x)=nx^(x-1)

So, in this case, you need to use power and chain rules.

f'(x)=(1/3)(x^2-3x-5)^(-2/3) (2x-3)

=(2x-3)/3(x^2-3x-5)^(2/3)


So, actually, you were close, you just missed bringing the 1/3 to the front.

Now see if you can get f''(x).


-Daniel-

 

[elindow]

So

f"=2-(2(2x-3)²)/(x²-3x-5)^1/3 ??

Thanks
Erik

 

[galactus]

The second derivative is a little trickier. Mostly, due to the algebra.

You can use the product rule:

\(\displaystyle \frac{2x-3}{3(x^{2}-3x-5)^{\frac{2}{3}}}\)

Product rule:

\(\displaystyle (\frac{2}{3}x-1)(\frac{-2}{3})(x^{2}-3x-5)^{\frac{-5}{3}}(2x-3)+(x^{2}-3x-5)^{\frac{-2}{3}}(\frac{2}{3})\)

\(\displaystyle =\frac{-2(2x-3)^{2}}{9(x^{2}-3x-5)^{\frac{5}{3}}}+\frac{2}{3(x^{2}-3x-5)^{\frac{2}{3}}}\)

Now, mulitply the right side, top and bottom, by \(\displaystyle 3(x^{2}-3x-5)\)

This will make the denominators the same and easier to work with.

\(\displaystyle \frac{-2(2x-3)^{2}}{9(x^{2}-3x-5)^{\frac{5}{3}}}+\frac{6(x^{2}-3x-5)}{9(x^{2}-3x-5)^{\frac{5}{3}}}\)

Adding and simplifying the numerators, you'll end up with:

\(\displaystyle \frac{-2(x^{2}-3x+24)}{9(x^{2}-3x-5)^{\frac{5}{3}}}\)

 

[elindow]

The second derivative is a little trickier. Mostly, due to the algebra.

You can use the product rule:

\(\displaystyle \frac{2x-3}{3(x^{2}-3x-5)^{\frac{2}{3}}}\)

Product rule:

\(\displaystyle (\frac{2}{3}x-1)(\frac{-2}{3})(x^{2}-3x-5)^{\frac{-5}{3}}(2x-3)+(x^{2}-3x-5)^{\frac{-2}{3}}(\frac{2}{3})\)

\(\displaystyle =\frac{-2(2x-3)^{2}}{9(x^{2}-3x-5)^{\frac{5}{3}}}+\frac{2}{3(x^{2}-3x-5)^{\frac{2}{3}}}\)

Now, mulitply the right side, top and bottom, by \(\displaystyle 3(x^{2}-3x-5)\)

This will make the denominators the same and easier to work with.

\(\displaystyle \frac{-2(2x-3)^{2}}{9(x^{2}-3x-5)^{\frac{5}{3}}}+\frac{6(x^{2}-3x-5)}{9(x^{2}-3x-5)^{\frac{5}{3}}}\)

Adding and simplifying the numerators, you'll end up with:

\(\displaystyle \frac{-2(x^{2}-3x+24)}{9(x^{2}-3x-5)^{\frac{5}{3}}}\)

Ok that doesn't make any sense to me if I use the quotiant rule I get:

f"(x)=(2x-3)/(3(x²-3x-5)^2/3)

= [3(x²-3x-5)^2/3(2)-(2x-3)[2(x²-3x-5)^-1/3(2x-3)]] / 3(x²-3x-5)^2/3²

which when I simplify is

[6(x²-3x-5)^2/3-2(2x-3)²(x²-3x-5)^-1/3] / 3(x²-3x-5)^2/3²

Cancel out the 3(x²-3x-5)^2/3 s
and I get

[2-2(2x-3)²] / 9(x²-3x-5)^2/3

 

[galactus]

OK then. We'll do it with the quotient rule:

\(\displaystyle \frac{2x-3}{3(x^{2}-3x-5)^{\frac{2}{3}}\)

Quotient rule:

\(\displaystyle \frac{3(x^{2}-3x-5)^{\frac{2}{3}}(2)-(2x-3)(2)(2x-3)(x^{2}-3x-5)^{\frac{-1}{3}}}{9(x^{2}-3x-5)^{\frac{4}{3}}}\)

Remember, squaring the denominator is part of the quotient rule.

=\(\displaystyle \frac{6(x^{2}-3x-5)^{\frac{2}{3}}-2(2x-3)^{2}(x^{2}-3x-5)^{\frac{-1}{3}}}{9(x^{2}-3x-5)^{\frac{4}{3}}}\)

=\(\displaystyle \frac{2}{3(x^{2}-3x-5)^{\frac{2}{3}}}-\frac{2(2x-3)^{2}}{9(x^{2}-3x-5)^{\frac{5}{3}}}\)

Multiply the left side, top and bottom, by \(\displaystyle 3(x^{2}-3x-5)\) to make the denominators the same. You'll get:

\(\displaystyle \frac{6(x^{2}-3x-5)}{9(x^{2}-3x-5)^{\frac{5}{3}}}-\frac{2(2x-3)^{2}}{9(x^{2}-3x-5)^{\frac{5}{3}}}\)

Simplify the numerator:

\(\displaystyle \frac{-2x^{2}+6x-48}{9(x^{2}-3x-5)^{\frac{5}{3}}}\)

Factor the numerator:

\(\displaystyle \frac{-2(x^{2}-3x+24)}{9(x^{2}-3x-5)^{\frac{5}{3}}}\)

There ya' go, the same thing as with the product rule.