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

Is there a faster way to find the derivative for this function?

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    Faster than what? You haven't proposed a solution or method. Also, please use [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). – 2017-02-07
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    How do you use mathjax? – 2017-02-07

3 Answers 3

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Find the derivative: $y=5(x+3)^2(2x-1)^5$

Although this particular problem is fairly straightforward using the product rule, when three or more factors are involved you might want to know about logarithmic differentiation.

Take $\ln$ of both sides to get

$$ \ln y=\ln5+2\ln(x+3)+5\ln(2x-1) $$

Take the derivative of both sides to get

$$ \frac{y^\prime}{y}=\frac{2}{x+3}+\frac{10}{2x-1}$$

Multiply on the left by $y$ and on the right by $5(x+3)^2(2x-1)^5$ to get

$$ y^\prime = \left(\frac{2}{x+3}+\frac{10}{2x-1}\right)\cdot5(x+3)^2(2x-1)^5$$

This may be taken further

$$ y^\prime = 10(x+3)(2x-1)^5+ 50(x+3)^2(2x-1)^4$$

factored

$$ y^\prime =10(x+3)(2x-1)^4[\,(2x-1)+5(x+3)\,] $$

and simplified to obtain a completely factor answer

$$ y^\prime = 10(x+3)(2x-1)^4(7x+14)$$

$$ y^\prime = 70(x+3)(2x-1)^4(x+2)$$

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    The only legitimately simple answer here. +1 – 2017-02-07
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    @BrevanEllefsen Thank you. – 2017-02-07
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$y'=10(x+3)(2x-1)^5+50(x+3)^2(2x-1)^4=70(x+3)(2x-1)^4(x+2)$

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    I try following the product rule, then I get lost in the middle of it. – 2017-02-07
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    I have trouble getting to that part 10(x+3)(2x-1)^5+50(x+3)^2(2x-1)^4 – 2017-02-07
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    @Fiona Lu It's just $(uv)'=u'v+uv'$ – 2017-02-07
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    Thanks for your help, I found the problem, I wasn't getting the right derivative for (2x-1)^5 – 2017-02-07
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    @Fiona Lu You are welcome! – 2017-02-07
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You can use product rule.

(uv) = uv' + u'v

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

$y=5\left[(x+3)^2(2x-1)^5\right]$

$y'=5\left[(x+3)^2 \cdot \frac{d}{dx}(2x-1)^5 + \frac{d}{dx}(x+3)^2 \cdot (2x-1)^5\right]$

$=5\left[(x+3)^2 \cdot 5(2x-1)^{5-1}\frac{d}{dx}(2x-1)+ 2(x+3)^{2-1}\frac{d}{dx}(x+3) \cdot (2x-1)^5\right]$

$=5\left[(x+3)^2 \cdot 5(2x-1)^4 \cdot 2+ 2(x+3)^1 \cdot 1 \cdot (2x-1)^5 \right]$

Hope you can proceed further.

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    Mine pleasure :-) – 2017-02-07