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I am having a Calculus test tomorrow ,and I've been practicing. I encountered this problem which I can't seem to find the solution: $ 5x(x-8)^{50}$. The textbook says that the answer is $5(x-8)^{49}(51x-8)$. I would like to know the way the book did it.

Thanks in advance.

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  • $\begingroup$ I am guessing you are not happy how they wrote $(x-8)^{50}$ as $(x-8)(x-8)^{49}$ and collect the term. $\endgroup$ – Chinny84 Mar 29 '17 at 22:41
  • $\begingroup$ $5x(x-8)^{50}$ can't be written as a composition of functions, so you can't just apply the chain rule to it. It can, however, be written as a product of functions... $\endgroup$ – Kaynex Mar 29 '17 at 22:46
  • $\begingroup$ @Kaynex One of the terms does require the use of the chain rule. (And the product rule is a special case of the chain rule, anyway.) $\endgroup$ – amd Mar 30 '17 at 1:20
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$${ \left( 5x(x-8)^{ 50 } \right) }^{ \prime }={ \left( 5x \right) }^{ \prime }{ \left( x-8 \right) }^{ 50 }+5x{ \left( { \left( x-8 \right) }^{ 50 } \right) }^{ \prime }=5{ \left( x-8 \right) }^{ 50 }+250x{ \left( x-8 \right) }^{ 49 }={ \left( x-8 \right) }^{ 49 }\left( 5x-40+250x \right) =\\ ={ \left( x-8 \right) }^{ 49 }\left( 255x-40 \right) =5\left( 51x-8 \right) { \left( x-8 \right) }^{ 49 }$$

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  • $\begingroup$ How did you get the 250x to add inside the $5(x-8)^{50}$? $\endgroup$ – Jfelix Mar 29 '17 at 22:54
  • $\begingroup$ which part is not clear to you? I think I have written quite clear $\endgroup$ – haqnatural Mar 29 '17 at 23:07
  • $\begingroup$ The transition from this $5(x−8)^{50}+250x(x−8)^{49}$ to this: $(x−8)^{49}(5x−40+250x)$ . $\endgroup$ – Jfelix Mar 29 '17 at 23:19
  • $\begingroup$ $(x-8)^49[5 (x-8)+250x]=(x-8)^49(5x-40+250x)$ $\endgroup$ – haqnatural Mar 29 '17 at 23:24
  • $\begingroup$ nevermind, you applied distributive property. I'm sorry. Thanks a lot for the solution! $\endgroup$ – Jfelix Mar 29 '17 at 23:35
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I am assuming you mean to differentiate $f(x) = 5x(x-8)^{50}$ with respect to $x$.

By the Product Rule:

$$ f'(x) = (\frac{d}{dx}(5x))\cdot (x-8)^{50} + (\frac{d}{dx}(x-8)^{50})\cdot 5x$$

By $\frac{d}{dx}(x^n) = nx^{n-1}$ :

$$ f'(x) = 5\cdot (x-8)^{50} + 50\cdot (x-8)^{49}\cdot 5x$$ $$ f'(x) = 5(x-8)^{49}((x-8)+50x)$$ $$ f'(x) = 5(x-8)^{49}(51x-8)$$

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A simple substitution will make differentiating this function easier. Let $u=(x-8)$. Then $x=(u+8)$ and

$$\\ 5x(x-8)^{50}=5(u+8)u^{50}=5(u^{51}+8u^{50})\\$$

Differentiating with respect to the original variable $x$ requires using the Chain Rule (though trivial in this case as $du = 1dx$):

$$\frac{d}{dx}5x(x-8)^{50} \\$$

$$=\frac{d}{dx}5(u^{51}+8u^{50}) \\$$

$$= 5\frac{d}{du}(u^{51}+8u^{50})\frac{du}{dx}\\$$

$$= 5(51u^{50}+8\cdot50u^{49})(1) \\$$

$$= 5u^{49}(51u+8\cdot50)$$

Re-inserting our original variable,

$$=5(x-8)^{49}(51(x-8)+50\cdot8) \\$$

$$=5(x-8)^{49}(51x-51\cdot8+50\cdot8) \\$$

$$=5(x-8)^{49}(51x-8)$$

Notice the pattern of the original problem, where simple factor $x$ had a nice power of 1, and the complicated factor $(x-8)$ had an undesirable power of 50:

$$5x(x-8)^{50}=5(u+8)u^{50}$$

Notice how our substitution $u=x-8$ gave us a simple factor of $u$ beneathe the power of 50, allowing us to avoid the product rule. If you like this answer, feel free to activate the check mark next to it.

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If you want to practice the chain rule:

$$\left[5x(x-8)^{50}\right]'=\left[5(x^{\frac{1}{50}+1}-8x^{\frac{1}{50}})^{50} \right]'=5\cdot50\cdot(x^{\frac{1}{50}+1}-8x^{\frac{1}{50}})^{49}\cdot(x^{\frac{1}{50}+1}-8x^{\frac{1}{50}})'=$$

$$250\cdot x^{\frac{49}{50}}(x-8)^{49}\cdot \left(\frac{51}{50}x^{\frac{1}{50}}-\frac{8}{50}x^{-\frac{49}{50}}\right)=250\cdot\frac{1}{50}\cdot(x-8)^{49}\cdot\left(51x^{\frac{49}{50}+\frac{1}{50}}-8x^{\frac{49}{50}-\frac{49}{50}}\right)=5(x-8)^{49}(51x-8).$$

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  • $\begingroup$ I think you mean a 5 and not 50 in the last line right after the equal sign, since 250/50=/=50. $\endgroup$ – smokeypeat Sep 16 '17 at 11:09
  • $\begingroup$ @user221227, thank you, corrected. $\endgroup$ – farruhota Sep 16 '17 at 12:43

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