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.

  • $\begingroup$ 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

$${ \left( 5x(x-8)^{ 50 } \right) }^{ \prime }={ \left( 5x \right) }^{ \prime }{ \left( x-8 \right) }^{ 50 }+5x{ \left( { \left( x-8 \right) }^{ 50 } \right) }^{ \prime }=5{ \left( x-8 \right) }^{ 50 }+250x{ \left( x-8 \right) }^{ 49 }={ \left( x-8 \right) }^{ 49 }\left( 5x-40+250x \right) =\\ ={ \left( x-8 \right) }^{ 49 }\left( 255x-40 \right) =5\left( 51x-8 \right) { \left( x-8 \right) }^{ 49 }$$

  • $\begingroup$ 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

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

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


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,

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

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


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:


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.


If you want to practice the chain rule:

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

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

  • $\begingroup$ 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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