# Thread: Help needed w/ differentiation

1. ## Help needed w/ differentiation

Hi, I am having trouble with a couple questions and my book does not feature any equivalent sample questions, nor does it explain how to deal with anything like them.

The problems are:

y = $\frac{\sqrt{x}-2}{\sqrt{x}+2}$

and

$F(x) = (x^5 - 1)^4 (2 - 3x)^3$

I have tried to apply the quotient, chain and product rules but haven't managed to get anywhere. If someone could explain the steps needed to solve these problems it would be much appreciated.

Thank you.

2. ## Re: Help needed w/ differentiation

For the first one, you can apply the quotient rule. You can use numeric exponents (instead of square roots). Before you differentiate:

$y = \dfrac{\sqrt{x}-2}{\sqrt{x}+2} = \dfrac{x^{1/2}-2}{x^{1/2}+2}$

Next, notice that if you add 2 and subtract 2 in the numerator, you get the denominator at least once:

$y = \dfrac{(x^{1/2}+2)-4}{x^{1/2}+2} = \dfrac{x^{1/2}+2}{x^{1/2}+2} - \dfrac{4}{x^{1/2}+2} = 1-4(x^{1/2}+2)^{-1}$

Do you know how to find the derivative of either of these two? I have not used any calculus yet, just algebra, so everything I just did should be straightforward.

For $F(x) = (x^5-1)^4(2-3x)^3$, you need to use the product rule and the chain rule multiple times. Let $g=(x^5-1)^4$ and $h=(2-3x)^3$. Then $F(x) = g\cdot h$. Taking the derivative using the product rule, you find $F'(x) = \dfrac{dg}{dx}h + g\dfrac{dh}{dx}$. To find $\dfrac{dg}{dx}$ and $\dfrac{dh}{dx}$, you can let $j = x^5-1$ and $k=2-3x$. This gives you $g=j^4$ and $h=k^3$. So, $\dfrac{dg}{dx} = 4j^3\dfrac{dj}{dx}$ and $\dfrac{dh}{dx} = 3k^2\dfrac{dk}{dx}$. You should be able to easily calculate both $\dfrac{dj}{dx}$ and $\dfrac{dk}{dx}$ and then just plug in what you find.

3. ## Re: Help needed w/ differentiation

Originally Posted by SDF
Hi, I am having trouble with a couple questions and my book does not feature any equivalent sample questions, nor does it explain how to deal with anything like them.

The problems are:

y = $\frac{\sqrt{x}-2}{\sqrt{x}+2}$

and

$F(x) = (x^5 - 1)^4 (2 - 3x)^3$

I have tried to apply the quotient, chain and product rules but haven't managed to get anywhere. If someone could explain the steps needed to solve these problems it would be much appreciated.

Thank you.
for (a) just use the quotient rule

$\dfrac{dy}{dx}=\dfrac{1}{\left(\sqrt{x}+2\right)^ 2}\left( \dfrac{d}{dx}(\sqrt{x}-2)\right)(\sqrt{x}+2)-(\sqrt{x}-2)\left(\dfrac{d}{dx}(\sqrt{x}+2)\right)$

(b) is a mess but just use the product and chain rule

$\dfrac{dF}{dx}=\left(\dfrac{d}{dx}(x^5-1)^4\right)(2-3x)^3 + (x^5-1)^4\left(\dfrac{d}{dx}(2-3x)^3\right)$

4. ## Re: Help needed w/ differentiation

SlipEternal,

For the second one, I got an answer of $(4x^5-4)^3(2-3x)^3 + (6-9x)^2(x^5-1)^4$ when I worked it out. I am not sure if this is correct, since I multiplied the inside terms with the exponent I "brought down" after applying the product rule. Could you verify this? If I'm wrong, I think it's because of that last step. in which case my answer would be $4(x^5-1)^3(2-3x)^3 + 3(2-3x)^2(x^5-1)^4$.

romsek,

I sort of understand what you did for the first one, apart from the first term. Why is it $\frac{1}{(\sqrt{x}+2)^2}$ and not $\frac{(\sqrt{x}+2)[\frac{d}{dx}(\sqrt{x}-1)] - (\sqrt{x}-1)[\frac{d}{dx}(\sqrt{x}+2)]} {(\sqrt{x}+2)^2}$

5. ## Re: Help needed w/ differentiation

Originally Posted by SDF
SlipEternal,

For the second one, I got an answer of $(4x^5-4)^3(2-3x)^3 + (6-9x)^2(x^5-1)^4$ when I worked it out. I am not sure if this is correct, since I multiplied the inside terms with the exponent I "brought down" after applying the product rule. Could you verify this? If I'm wrong, I think it's because of that last step. in which case my answer would be $4(x^5-1)^3(2-3x)^3 + 3(2-3x)^2(x^5-1)^4$.
\displaystyle \begin{align*}F'(x) & = (x^5-1)^4\left[{\dfrac{d}{dx}\left(2-3x\right)^3}\right] + (2-3x)^3\left[{\dfrac{d}{dx}\left(x^5-1\right)^4}\right] \\ & = 3(x^5-1)^4(2-3x)^2 \left[{\dfrac{d}{dx}(2-3x)}\right] + 4(2-3x)^3(x^5-1)^3 \left[{\dfrac{d}{dx}(x^5-1)}\right] \\ & = -9(x^5-1)^4(2-3x)^2 + 20x^4(2-3x)^3(x^5-1)^3 \\ & = (x^5-1)^3(2-3x)^2(-9x^5+9+40x^4-60x^5) \\ & = (x^5-1)^3(2-3x)^2(9+40x^4-69x^5)\end{align*}

That's about as simplified as I can get it. You don't need to do any of that algebra. You can stop at $F'(x) = -9(x^5-1)^4(2-3x)^2 + 20x^4(2-3x)^3(x^5-1)^3$.

Note: $-9(x^5-1)^4 \neq (-9x^5+9)^4$ This is because $a^cb^c = (ab)^c$. So, in order to bring the $-9$ into the parentheses, you would need to take the fourth root. But, you cannot take an even root of a negative number over the reals. So, you could write $-9(x^5-1)^4 = -(\sqrt[4]{9}(x^5-1))^4$ if you wanted, but it is certainly not necessary.

6. ## Re: Help needed w/ differentiation

Originally Posted by SDF
I sort of understand what you did for the first one, apart from the first term. Why is it $\frac{1}{(\sqrt{x}+2)^2}$ and not $\frac{(\sqrt{x}+2)[\frac{d}{dx}(\sqrt{x}-1)] - (\sqrt{x}-1)[\frac{d}{dx}(\sqrt{x}+2)]} {(\sqrt{x}+2)^2}$
You seem to be missing something.

$\dfrac{(\sqrt{x}+2)[\tfrac{d}{dx}(\sqrt{x}-2)]-(\sqrt{x}-2)[\tfrac{d}{dx}(\sqrt{x}+2)]}{(\sqrt{x}+2)^2} = \dfrac{1}{(\sqrt{x}+2)^2}\left((\sqrt{x}+2)[\tfrac{d}{dx}(\sqrt{x}-2)]-(\sqrt{x}-2)[\tfrac{d}{dx}(\sqrt{x}+2)]\right)$

Division is the same as multiplying by a reciprocal. romsek just wrote the quotient rule as the multiplication of the reciprocal of the square of the denominator and the same numerator that you had.