# How to find the derivative of the function $f(x) = (2x-3)^4 (x^2 + x + 1)^5$?

Can someone please tell me how to find the derivative of this function? I've been working on this one problem since yesterday and I still can't find the answer...

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

apparently you must use the product rule and the chain rule but I'm totally confused on how to get the answer...

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The exclamation points deter me from wanting to answer. I agree - you just use the product rule and the chain rule. What did you try, what did you get, and why do you think it's incorrect? –  mixedmath Feb 20 '13 at 1:39
Desperate, and the doubt is purely algorithmical. You surely didn't try hard enough. Sorry, but had to downvote. –  Aloizio Macedo Feb 20 '13 at 1:43
Do you know about the product and the chain (or power) rule? –  ncmathsadist Feb 20 '13 at 1:45
You're offended by exclamation marks?? Wow. And I am doubtful the answer is correct due to the simple fact that my answer does not match the answer in the back of the book. Now to address exclamation marks..these are to show emotion in the fact that I need help to find the answer before the homework is to be turned in tomorrow. –  bkaifos15 Feb 20 '13 at 1:46
@bkaifos15: What is your answer, and (more importantly) how did you arrive at it? –  Blue Feb 20 '13 at 1:47

You want the derivative of $f(x)= (2x-3)^4 (x^2+x+1)^5$. Take it one step at a time. If you were evaluating $f(x)$ for a particular value of $x$, the last calculation that you’d make is multiplying $(2x-3)^4$ and $(x^2+x+1)^5$, so that product is the first thing to take care of in the differentiation. Apply the product rule:

$$f\,'(x)=(2x-3)^4\left[(x^2+x+1)^5\right]'+\left[(2x-3)^4\right]'(x^2+x+1)^5\;.$$

Now you have to take the derivatives $\left[(x^2+x+1)^5\right]'$ and $\left[(2x-3)^4\right]'$. Each of these is the derivative of a power, so use the power rule. But they are not powers of $x$, so you’ll also need the chain rule:

$$\left[(x^2+x+1)^5\right]'=5(x^2+x+1)^4\left[x^2+x+1\right]'\;,$$

and

$$\left[(2x-3)^4\right]'=4(2x-3)^3\left[2x-4\right]'\;.$$

All that’s left is to differentiate the polynomials $x^2+x+1$ and $2x-4$, getting $2x+1$ and $2$, respectively. Now put the pieces back together:

\begin{align*} f\,'(x)&=(2x-3)^4\left[(x^2+x+1)^5\right]'+\left[(2x-3)^4\right]'(x^2+x+1)^5\\ &=(2x-3)^4\Big(5(x^2+x+1)^4(2x+1)\Big)+\Big(4(2x-3)^3(2)\Big)(x^2+x+1)^5\\ &=5(2x+1)(2x-3)^4(x^2+x+1)^4+8(2x-3)^3(x^2+x+1)^5\;. \end{align*}

That’s a perfectly reasonable answer, but you could also pull out a bunch of common factors and write it as

\begin{align*} f\,'(x)&=(2x-3)^3(x^2+x+1)^4\Big(5(2x+1)(2x-3)+8(x^2+x+1)\Big)\\ &=(2x-3)^3(x^2+x+1)^4(28x^2-12x-7)\;, \end{align*}

which many would prefer.

The key is to take it one step at a time, working backwards through the calculation of $f$: here $f$ is a product at the outermost level, so the first step in the differentiation uses the product rule.

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Thank you so much I really appreciate your help!!! And you not being offended by my !!!!! marks and CAPITAL LETTERS! :] Thank you so much again! –  bkaifos15 Feb 20 '13 at 1:52
@bkaifos15: You’re welcome. (But it is a good idea to avoid ALL-CAPS: most people will interpret it as the visual indication of shouting.) –  Brian M. Scott Feb 20 '13 at 1:55
Ok note taken haha :P Definitely my first time using this site so didn't know there were rules on how to speak without offending :P –  bkaifos15 Feb 20 '13 at 1:57

Here is how you can find derivatives of similar expression:

Note that if you are looking for $f'(x)$, then $$(\ln f(x))'=\frac{f'(x)}{f(x)},$$ hence $$f'(x)=f(x)(\ln f(x))'.$$ Now we only need to find $(\ln f(x))'$, but this is simple matter, since $$\ln f(x)=4\ln(2x-3)+5\ln(x^2+x+1).$$ Therefore, $$(\ln f(x))'=\frac{8}{2x-3}+\frac{5(2x+1)}{x^2+x+1},$$ hence the final answer is $$f'(x)=\left(\frac{8}{2x-3}+\frac{5(2x+1)}{x^2+x+1}\right)f(x),$$ which is exactly what @BrianM.Scott obtained in his answer. The advantage of this method is that you can deal with much more complicated expressions.

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(2x-3)^4 (x^2+x+1)^5 first you have to take the derivative (2x-3)^4 and (x^2+x+1)^5 dy/dx=(2x-3)^4 is 4(2x-3)^3 [(2)] dy/dx=(x^2+x+1)^5 is 5(x^2+x+1)^4[(2x+1)] dy/dx=(2x-3)^4[5(x^2+x+1)^4(2x+1)]+4(2x-3)^3 (2)^5 dy/dx=5(2x+1)(2x-3)^4(x^2+x+1)^4 + 8(2x-3)^3)(x^2+x+1)^5

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Please read the editing help on how to format your posts. In particular, look up how to format equations using TeX for better readability. –  Mårten W Sep 9 '13 at 0:03

With the definition as an aid, make sense of the following limit expression by identifying the function $f$ and the number $a$. $$f′(\pi)=\lim_{h→0}\frac{\sin(\pi+h)−\sin(\pi)}{h}$$ is the derivative of the function $f(x)=$ at the number $a$.

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This doesn't relate to the question at all. There are no sines in the question and you mention nothing about the product or chain rules. I will fix your latex, but it would be good to see the LaTeX tutorial, as well. –  robjohn Nov 21 '13 at 18:50