# Math Help - Derivative Help using Quotient & Chain Rule

1. ## Derivative Help using Quotient & Chain Rule

Can someone please help me out... I've been struggling with this question for a while. I'm decent at doing just the chain rule or just the quotient rule but for some reason I can't do them together. The question is:

Find the derivative of:

1/ square root of 2x-3

One part that really confuses me is when i do the quotient rule:

f' (x) g (x) - g' (x) f (x) = 0 - 1x-3^-1/2 (1) / g(x)^2 = ???

I can't seem to figure out g(x) squared... G (x) is 2x-3^-1/2, and if we do that squared would it be just 2x-3 ?? (1/2 x 2 = 1) ?

if that's the case I have something like:

-x-3^-1/2 / (2x-3) squared

= square root (-x-3) / (2x-3) squared...

...

as you can see it's a jumbled mess.. If someone can provide me an answer step by step that'll really help. I tried it but can't seem to get it. Thanks in advance!

2. I would recommend re-writing the question as:

$\left(2x - 3\right)^{-\frac{1}{2}}$

Then, the derivative is found by applying the Power Rule:

$(-\frac{1}{2})\left(2x - 3\right)^{-\frac{1}{2} - 1}
\times \dfrac{d}{dx}\left[2x-3\right] = (-\frac{1}{2})\left(2x - 3\right)^{-\frac{3}{2}} \times 2 = -\left(2x - 3\right)^{-\frac{3}{2}}$

Of course, you can re-write the final answer as:

$-\dfrac{1}{\left(2x-3\right)^{\frac{3}{2}}}$

Using the Quotient Rule, let $f(x) = 1$ and $g(x) = \sqrt{2x-3}$. Then, $f'(x) = 0$ and $g'(x) = \dfrac{1}{\sqrt{2x-3}}$.

The derivative is given by: $\dfrac{f'(x) \times g(x) - f(x) \times g'(x)}{[g(x)]^2}$.

Substitute the values of f(x), f'(x), g(x), and g'(x) into the above equation.

$\dfrac{0 \times \sqrt{2x - 3} - 1 \times \dfrac{1}{\sqrt{2x-3}}}{(\sqrt{2x-3})^2} = \dfrac{-\dfrac{1}{\sqrt{2x-3}}}{2x - 3} = -\dfrac{1}{(2x-3)^{\frac{3}{2}}}$

3. Thanks so much... it makes so much more sense doing it the power rule way... I understood steps of it when I did it but messed up a lot... I still don't fully understand it if we went through the quotient way because I don't see where the 3/2 comes from since it's -1 / square root 2x -3 / 2x - 3... I understand that whole part, just not how to do it further...

The test I have coming up says it's going to be the chain rule along with other stuff... do they usually specify whether we have to use quotient rule with the chain rule or is it up to us to solve it the way we want to?

Thanks for the help

Also what happens with the 1 at the top? we turn the equation into 2x-3^-1/2 but what happens to the 1 that was given to us? it practically disappeared... would that always happen for the numerator? What happens if it's a number other than 1 such as:

5/ square root of 2x-3

4. Originally Posted by agent2421
Thanks so much... it makes so much more sense doing it the power rule way... I understood steps of it when I did it but messed up a lot... I still don't fully understand it if we went through the quotient way because I don't see where the 3/2 comes from since it's -1 / square root 2x -3 / 2x - 3... I understand that whole part, just not how to do it further...

The test I have coming up says it's going to be the chain rule along with other stuff... do they usually specify whether we have to use quotient rule with the chain rule or is it up to us to solve it the way we want to?

Thanks for the help

Also what happens with the 1 at the top? we turn the equation into 2x-3^-1/2 but what happens to the 1 that was given to us? it practically disappeared... would that always happen for the numerator? What happens if it's a number other than 1 such as:

5/ square root of 2x-3

What happens is you simply multiply it. You do the same with the one, it's just understood anything multiplied by 1 is itself. This concept had me for a while when I was learning chain rule because most first year calculus teachers assume their students operate at a much higher level of mathematical thinking than they actually do. This is good if the student wants to be mathematically strong, like myself, but bad for students that only need first year calculus - like business/econ majors. But I digress....

Here's what the work actually looks like.

$\frac{5}{\sqrt{2x-3}}$

is the same as

$5*(2x-3)^\frac{-1}{2}$

Which is the same as

$5 * \frac{1}{\sqrt{2x-3}}$

Even though this is a relatively simple topic, it's crucial to understand in order to know when to use the chain rule to simplify vs. using the quotient rule. Whenever a constant is in the numerator it's often easier to use the chain rule than the quotient rule.

Does this make sense?

$\dfrac{-\dfrac{1}{\sqrt{2x-3}}}{2x - 3}$

Division by a number is the same as multiplying by the reciprocal of that number. In this case, division by $2x - 3$ is the same as multiplying by the reciprocal of $2x - 3$, which is $\dfrac{1}{2x-3}$.

$\dfrac{-\dfrac{1}{\sqrt{2x-3}}}{2x - 3} = -\dfrac{1}{\sqrt{2x-3}} \times \dfrac{1}{2x - 3}$

Multiply the numerators and multiply the denominators to combine the fractions.

$-\dfrac{1}{\sqrt{2x-3}} \times \dfrac{1}{2x - 3} = -\dfrac{1 \times 1}{\sqrt{2x-3} \times (2x-3)} = -\dfrac{1}{(2x-3)\sqrt{2x-3}}$

Now, re-write $\sqrt{2x-3}$ as $(2x-3)^{\frac{1}{2}}$.

$-\dfrac{1}{(2x-3)\sqrt{2x-3}} = -\dfrac{1}{(2x-3)(2x-3)^{\frac{1}{2}}}$

Finally, $(2x-3)$ and $(2x-3)^\frac{1}{2}$ have a common base, so add exponents to find their product. (The exponent on $(2x-3)$ is implied to be 1.)

$-\dfrac{1}{(2x-3)(2x-3)^{\frac{1}{2}}} = -\dfrac{1}{(2x-3)^{1 + \frac{1}{2}}} = -\dfrac{1}{(2x-3)^{\frac{3}{2}}}$

$-\dfrac{1}{(2x-3)^{\frac{3}{2}}}$

6. Originally Posted by agent2421
Can someone please help me out... I've been struggling with this question for a while. I'm decent at doing just the chain rule or just the quotient rule but for some reason I can't do them together. The question is:

Find the derivative of:

1/ square root of 2x-3

One part that really confuses me is when i do the quotient rule:

f' (x) g (x) - g' (x) f (x) = 0 - 1x-3^-1/2 (1) / g(x)^2 = ???

I can't seem to figure out g(x) squared... G (x) is 2x-3^-1/2, and if we do that squared would it be just 2x-3 ?? (1/2 x 2 = 1) ?

if that's the case I have something like:

-x-3^-1/2 / (2x-3) squared

= square root (-x-3) / (2x-3) squared...

...

as you can see it's a jumbled mess.. If someone can provide me an answer step by step that'll really help. I tried it but can't seem to get it. Thanks in advance!

$\displaystyle\ h(x)=\frac{f(x)}{g(x)}$

$\displaystyle\ h'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

$f(x)=1\Rightarrow\ f'(x)=0$

$g(x)=\sqrt{2x-3}=(2x-3)^{0.5}=u^{0.5}$

$\displaystyle\ g'(x)=0.5u^{-0.5}u'(x)=\frac{2}{2\sqrt{u}}=\frac{1}{\sqrt{2x-3}}$

Just apply the Quotient Rule and use the Chain Rule independently for the part that requires it.

$\displaystyle\ h'(x)=\frac{0-\frac{1}{\sqrt{2x-3}}}{2x-3}=-\sqrt{2x-3}^{-3}$

Of course, you could simply use

$h(x)=(2x-3)^{-0.5}$

and bypass the Quotient Rule.